Warsha (3)
155 Question
On
Chapter 1
1
Mr. Hytham Ahmed
2
Mr. Hytham Ahmed
3
Mr. Hytham Ahmed
Brightnesssssssssss
P=
Brightness = Power
1st case series connection only
ππΏπππ 2
π
πΏπππ
V
If (A) burned what happened to brightness (B)
Reading of voltmeter (Inc – dec – same)
Ans
A
β΅ π πΌ π 2 ππππ¦
β΅ Rlamp Constant
If V ↑ → P ↑
B
&
π ↓∴π↓
2nd Case parallel only
β΅r=0
What happen if lamp (A) burned to
1- Brightness of (B) → Same (
ππ΅2
π
2- To reading of A → Same ( πΌ =
)
ππ΅
π
1
)
3- To reading of V → Same (V = VB)
4
Mr. Hytham Ahmed
3rd Case parallel & Series
If (B) burned what happen to brightness of (A) & (C)
VB = Vs + Vparallel
B
RParallel ↑ → VParallel ↑
↑Pc ο ↑ VC
A
VSeries ↓ β΅ VA ↓ → PA↓
C
What happen to the brightness?
1-in the following circuit: what's happen to
brightness of lamps when key is closed?
Req parallel dec then Vparallel ( decreases ) then brightness decrease
At same time VSERIES ( increase ) then brightness increase
∴ X1 Bright,
X2 Decrease
, X3 increase
2-WHAT happen to brightness of lamp if key is closed
-It will put off
5
as wire draw all current
Mr. Hytham Ahmed
3- What happen to brightness of each lamp if key k is closed
Lamp 1 : Remain Constant As Voltage is constant
Lamp 2 : Increase As its Voltage increase
Lamp 3 : Put off
4- What happen to brightness of lamp if rheostat increases?
Remain Unchanged As its Voltage Unchanged
Remain unchanged as it's voltage unchanged so π =
π2
π
It's constant
6
Mr. Hytham Ahmed
1)
1-calculate I in 6Ω
2- power consumed at 2Ω
π
18
π΅
Sol: 1) I = π
+π
= 5+1 = 3A
2) Simplify the circuit to one branch
1) Vt = I x R = 3 x 2 = 6V
2) Vt = V1 = V2 = 6V
3) I1 = 6⁄3 = 2A
4) I2 = 6⁄6 = 1
5) Take the ( 2I ) to the original circuit
6) Take the this part out + 3steps
Vt = 1 x 4 = 4V
I1 = 4⁄6 = 2⁄3 A
I2 =4⁄12 = 1⁄3 A
I in 6 Ω = 1/3 A
P=V x I = I2 x R =
π£2
π
P = I2 x R = 12 x 2 = 2 watt
7
Mr. Hytham Ahmed
2) find R
Sol: -
r=0
Ir = 0 Vex = VB = 1.5V
V10Ω = VR = 0.003 x 10 =0.3V
V8Ω = 1.5 – 0.3 = 1.2V
π
1.2
π
8
I in 8 Ω = =
= 0.15π΄
0.15 = total current
I in R = 0.15 – 0.003 = 0.12
V = 0.3V
At R
I = 0.12 A
π
0.3
πΌ
0.12
R= =
= 2.5 Ω
Note: - take care r = 0 then Vex = VB without thinkiiiiing
8
Mr. Hytham Ahmed
3) Find R
Sol: - R = 4 Ω
4) Find R1 &R2
Sol: - R1 = 6 Ω, R2 = 4 Ω
9
Mr. Hytham Ahmed
5) Find: -
1) VB
2) r
Sol: - relation
V= VB – Ir
VB = 12V from the fig.
Slope = r
Slope =
π¦2 −π¦1
π₯2− π₯1
π½
π½
π°
π°
6) Note: •If =
=
12−8
4−0
=1
it means R is const then R obey ohm’s law.
A student find a device as Resistor, V = 1.5 V, I = 45x10-6 A But when 3 volt
Battery used I=25x10-3A .
Dose the Device obey ohm’s law? why?
Sol: Not obey as V/I in the first case not equals that in second case
10
Mr. Hytham Ahmed
7)
1) calculate A1 , V1 ,V2 if the key is open and if it
is closed
2) Ι³ of Battery, % drop in voltage
3) what’s happen to A1 , V1 ,V2 if rheostat
increases
Sol: - 1) when key is open, I = 0 , A= 0 , V1 = 0 , V2 = VB =30V
• Close I =
30
8+2
= 2A , A = 3A
V1 = I x R = 3 x 6 = 18V
V2 = VB – Ir = 30 – 3 x2 = 24V
2) Ι³ =
πΌπ
ππ΅
x 100 =
% drop =
πΌπ
ππ΅
(3)(8)
30
x 100 =
x 100 = 80%
(3)(2)
30
= 20%
3) if the rheostat increases Rt inc and I dec therefore A dec
V1 = IR
then V1 Dec
V2 = VB – Ir then V2 inc
11
Mr. Hytham Ahmed
8) Find VB if power consumed at 2Ω = 8 watt
Sol: - any prop contain battery we use this
I=
ππ΅
I=
π
+π
P = I2 R =
ππ΅
4+1
π2
π
= VI
8 = I2(2) = I2 = 4
I = 2A this is not It
V = 12V
π
12
π
3
I3Ω = =
= 4A
It = 4+2 = 6A
Then 6 =
ππ΅
5
VB = 30V
9) Find VB, r if (A) reads 2A when K is open and read
3A when k is closed.
•If he wants VB, r
we make two equation and solve it together
Open I =
ππ΅
π
+π
Closed 3 =
ππ΅
6+π
, 2=
ππ΅
10+π
(1)
(2)
2(10 + r ) = 3(6 + r)
20 + 2r = 18 + 3r
r = 2Ω
VB = 24Ω
12
Mr. Hytham Ahmed
10) When the battery connected to a wire 6m length R = 12Ω and I pass = 0.3 A
Then when wire Decreased to 2m and connected to same battery I becomes
0.5A.
-Find emf and the internal Resistance of Battery then if A = 0.24mm2, find
Resistivity of material of wire.
Sol: - VB = 6Ω, r = 8 Ω, Pe = 4.8x10-7 Ω.m
11) A wire stretched uniformly and its radius Dec to third. Calculate
πΉπ
πΉπ
{stretched, compressed, reformed, reshaped, Drawn}
r1 = 3r2 (squaring) → A1 = 9 A2 (just replace the number) → 9L1 = L2
π
1 πΏ1 π΄2 ππ1
=
π
2 πΏ2 π΄1 ππ2
π
1
π
2
=
πΏ1 π΄2 ππ1
9πΏ1 9π΄2 ππ1
=
1
81
If it was one wire, then the length and the area must be changed inversely
Note {stretched uniformly, wrapped, reformed} the both L, A must be changed
inversely
13
Mr. Hytham Ahmed
12) A wire stretched uniformly and if the length increased 3 times find
πΉπ
πΉπ
Sol: - {1/9}
13) two wires of same material length of the 1st 2 times that of 2nd Find
πΉπ
πΉπ
Sol: - L1 = 2L2
He didn’t say anything about area then A1=A2
π
1
π
2
=
πΏ1
πΏ2
same material Pe1=Pe2
π
1
2πΏ2
2
=
=
π
πΏ2
1
14
Mr. Hytham Ahmed
14) 2wires of same material L1 = 0.1m, L2= 0.5m, m1 = 0.2kg, m2 = 0.1kg
-Find R1/R2
•Mass problem:
π
1
π
2
=
π
1
π
2
=
πΏ21 π2
πΏ22 π1
(0.1)2 (0.1)
(0.5)2 (0.2)
15) Find reading of V when the key is open and when it
closed
Sol: -Open I =
ππ΅
π
+π
=
90
45+0
= 2A
V= IR = 2 x 30 = 60V
-Closed I =
ππ΅
π
+π
=
90
30+0
= 3A
V = IR = 3 x 30 = 90V
15
Mr. Hytham Ahmed
16) Find A1, V. When key is open and when it is closed
Sol: Open A=3A V=9V
Closed A=0A V= 18V
17) 3 Resistors 10, 20, 30 current passing through there 0.15A,0.2A,0.5
respectively
-Find equivalent R with draw.
Sol:
•R1 = 10
I1 = 0.15A
v1= 1.5V
•R2=20
I2 = 0.2A
V2=4V
•R3 = 30
I3 = 0.05A
V3=1.5V
Bec of voltage equal then they are parallel
RT=27.5Ω
16
Mr. Hytham Ahmed
19) 3 Resistors 20,40,60Ω connected together to battery pot. Diff across each R
is 50,20,30V respectively
Find equivalent R with Draw.
RT=50/3 Ω
20) -Find V when
•K1, K2 closed
•K1, K2 open
•K1 closed K2 open
•K1 open K2 closed
Sol: 1)I=
12
4+2
= 2A , V = 2 x 4 = 8V
2)V = VB = 12V
3)I =
4)I =
17
12
8+2
12
8+2
= 1.2A, V= 1.2 x 8 =9.6V
= 1.2A, V= 1.2 x 8 =9.6V
Mr. Hytham Ahmed
21) Find (V)
•K1 , K2 , K3 closed
•K1 , K2 , K3 open
•K3 open , K1 , K2, closed
22)
1) Find I in 12-volt battery
2) power consumed at 24Ω
Sol: Look at I in each battery if they are in the same direction then VB1+VB2 if not then VB1-VB2
I=
ππ΅1−ππ΅2
π
+π+π
12−6
= 18+3+3 = ¼ A
Move with I from the bigger battery
Vt = ¼ x 8 = 2V
I in 24Ω = 1/12 A
P= I2 R = (1/12)2 x 24 = 1/6 watt
Note if he want (v) of
-Bigger battery → V= VB – Ir
-Smaller battery → V=VB + Ir
18
Mr. Hytham Ahmed
23) An electric circuit of cell (2v, 0Ω) connected in series with 2 Resistance AB =
10Ω Bc= 10Ω , then 2 other Resistance AD=4Ω , DC = 6Ω were parallel with the
previous Resistance
-Calculate: •Pot Diff between A,C
•Pot Diff between B,D
Sol:r=0, Ir=0, Vex=VB → VAC =2V
• VBD → He wants pot. diff. between 2 points each one has I →
VBD = VAB – VAD
2
I = 20 π₯ 10
20+10
+0
= 0.3 A
V = IR = 0.3 x
200
30
VBD = IR – IR = (0.1)(0.1) – (0.2) (4) = 0.2 Volt
19
Mr. Hytham Ahmed
24) if electric generator is connected to factory by 2 wires of length 1Km each.
V at generator 230V, at factory is 230 V, the current intensity in wire = 50A, if Pe
of the wire =1.57 x 10-6 Ω.m
Find:
1)the R per meter
2) and radius of wire
3) percentage of Ι³ of transportation
Sol: V = 230 – 220 = 10V, I= 50A → R = 0.2Ω
1)R/meter =
2) Pe =
π
π΄
πΏ
0.2
1000 π₯ 2
→ 1.57 x 10 -6 =
0.2 π΄
1000 π₯ 2
π΄ = ππ 2 , r= m →(2)
3) Ι³ =
ππππππ£ππ
ππππππ’πππ
x 100
Pproduced = VG x I = 230 x 50
Plost = I2 R = 502 x 0.2
ππππππ£ππ = Pproduced - Plost
20
Mr. Hytham Ahmed
25) In the circuit when the key is closed the total resistance
decreased to half its value
Calculate the value of ( R ) knowing that r=0
Sol:Key open
RT1 = 50 + R
Key closed
RT2 = 18 + R
RT2 = ½ RT1
(1)
(2)
given
18 + R = ½ ( 50 + R )
18 + R = 25 + ½ R
½R=7
R = 14Ω
Note :- any empty wire its two ends take the same number → if the I is divided and
didn’t get back again.
26) Find Rt
Sol :- 6Ω
21
Mr. Hytham Ahmed
27) Find Rt
Sol :- 6Ω
28) Find
1) V at B
2) V at C
3) internal Resistance
Sol: Vab = I R = Va – Vb
if he gives me V at point
VAB = IR = VA - VB
10 x 2 = 50 – V (at B)
V(at B) = 30 Volt
VCD = IR = Vc – VD
Vc – 0 = 10 x ½
VC = 5V
VBC = VB – VC = VB + Ir
(+) Bec the arrow direction is to the battery
30 – 5 = 10 + 10 r
r = 1.5 Ω
22
Mr. Hytham Ahmed
29) Find V between A,B
Sol:V6Ω = I x R = 4 x 6 = 24 V
V of 12Ω = 24 V Bec they are //
π
24
π
12
I in 12 Ω = =
= 6A
I in 2 Ω = 2 + 4 = 6A
V( 2 Ω + ( 6 // 12) = 6x6 = 36
V 3 Ω = 36
I in 3 Ω =
36
3
Bec they are //
= 12 A
It = 12 + 6 = 18A
VAB = I x R = 18 x 5 = 90 V
30) Find VB of battery
VB = 22V
23
Mr. Hytham Ahmed
31) find VB
VB = 24
32) A wire carry a I = 0.1 A and V= 1.2V reformed as square ABCD
Calculate Total Resistance if Battery connect between
1) AD
2) AC
Sol :π
1.2
πΌ
0.1
R= =
3 // 9 Rt =
= 12Ω
3π₯9
3+9
= 2.25Ω
6//6 Rt = 3 Ω
24
Mr. Hytham Ahmed
33) Find R
Note: - if two currents equal and opposite in direction enter to a resistance it is cancelled
↓
25
Mr. Hytham Ahmed
34) if each resistance = 6 Ω , find:
1) equivalent resistance
2) the reading of Ammeter A1, A2
1)4 Ω
2) A1 = 0, A2 = 3A
35) what happen to reading of voltmeter when rheostat increases + r not equal
zero?
R inc → π° dec
1-
2-
3-
4-
26
Mr. Hytham Ahmed
36) A current of 2 A pass in a ring of radius 8cm and the
potential difference between its terminals Equal (2π
) Volt.
and cross section area of the wire of the ring = 0.2 cm2.
Calculate: The resistance of the wire if the ring electric
resistivity SAME the wire material.
Sol:
π½
ππ
π°
π
Req = =
2
π
=π
π
Req= = π
R = Req x 2 = 2π
2
2
π
∴ π
ππ ππ π‘ππππ ππ π€πππ = π
+ π
= 2π + 2π = 4π Ω
37) What is the value of RV that makes
consumed power at R1=36 WATT
P= I2 R
36= I2x9
I1R1=I2R2
I=2 A
2x9=18xI2
I2 = 1 A
I=I1 + I2 = 3A
πΌ=
ππ΅
π
ππ+π
Req = RV+ R//
27
3=
39
π
ππ+2
Req = 11 β¦
RV =11 - 6 = 5β¦
Mr. Hytham Ahmed
38) from study the following figure that illustrate the relation between potential
dropped and points on circuit find reading of ammeter and value of resistance
As V(D-E)=IR1=1 x I=(12-10)"from graph"
So I=2 (A)
,
As V(R)=V(G-F)-IR3=10-3 x 2=4
So Req(parallel)= V/I = 4/2 = 2 ohm ,
28
πππΉ
π+πΉ
=2
R= 6β¦
Mr. Hytham Ahmed
39) Three identical resistors are connected as
shown to battery of:
1) what change in current intensity through the
electric cell happens when the resistor X is
replaced by a wire of zero resistance
2) find the ratio between the readings of
ammeter A before and after replacing resistor X
Sol: (a) Resistance decrease (y only ) so, current increase
πΌ
(b) before , Ibef= π‘ππ‘ πΌπ‘ππ‘ =
2
After, IBEFORE= ITOT / 2
Iafter =
29
ππ΅
π
∴
πΌπππ
πΌπππ‘ππ
=
ππ΅
π
+0.5π
=
ππ΅
3π
π₯
=
2ππ΅
3π
ππ΅
3π
π
ππ΅
=
1
3
Mr. Hytham Ahmed
40) An electric circuit contains four electric resistors ( R1,R2,R3,R4)ohm , If the
electric current intensity that flows through each of them is ( 0.3,0.3, 0.4, 0.2)
ampere respectively and the value of R1 =6β¦, R3= 15β¦ And the internal resistance
of the battery 1β¦ Show by drawing the method of connection of these resistors
.Calculate the total resistance of the circuit .Calculate the electromotive force of
the electric source. (14β¦,8.4V)
OR (7.6β¦, 6.9V)
ANS:
b) V1 = V2
,
0.3 x 6 = 0.3 R2 → R2 = 6 ohm
V3 = V4
,
0.4 x 15 = 0.2 R4 → R4 = 30 ohm
6
15 π₯ 30
2
15+30
Rt = +
+ 1 = 14 ohm
c) I = 0.3 + 0.3 = 0.6 A
VB = I ( R+r ) = 0.6 x 14 = 8.4 V
30
Mr. Hytham Ahmed
41) In the given electric circuit, what would happen
to the reading of both voltmeters (V1) and (V2) when
increasing the value of the variable resistance (R1)?
as R1 inc ... current in circuit decrease
So V2 = I R2 (decrease) and V1= IR1 (increase)
42) When key closed, in the opposite figure:
a) Ammeter reading (A) …..
( decrease – increase – not change )
b) Voltmeter reading (V1) …..
( decrease – increase – not change )
c) Voltmeter reading (V2) …..
( decrease – increase – not change )
31
Mr. Hytham Ahmed
43) which lamp must cut off to obtain max
reading of voltmeter
( A –B –C )
Answer:
Lamp B β due to as B burn I =0 and then V=VB (max value)
32
Mr. Hytham Ahmed
44) If reading of ammeter when key is
open = 2 A ,
and when closed reading becomes
2.4 A , then
internal resistance equal……
βΆ 1Ω
β· 2Ω
K is opened
I=
ππ΅
10+π
=2A
βΈ 0.5Ω
,
ππ΅
8+π
= 2.4 A
12// 12 = 6 ,
20 + 2r = VB
K is closed
I=
βΉ 1.5Ω
,
,
Req = 4 +6 = 10Ω
→(1)
12// 6 = 4 ,
19.2 + 2.4 r = VB
Req = 4 +4 = 8Ω
→ (2)
From (1&2)
20 + 2r =19.2 + 2.4 r
33
∴r=2Ω
Mr. Hytham Ahmed
45) Two wires (ab) and (cd) of the same
material having the same the length are
connected together in series in a closed
circuit. If the wire (ab) is thicker than the
wire (cd), which potential difference is
greater, that across the wire (ab) or
across the wire (cd)? Explain your
answer.
Answer:
The voltage across (cd) is greater.
R=πe
Rα
1
π΄
πΏ
same length and same material
π΄
β΅ Acd< π΄ππ
∴ Rcd>Rab
β΅ Vα R at constant I
∴ Vcd>Vab
34
Mr. Hytham Ahmed
46) Find the value of I1 &I2 & I3 when the Reading of the voltammeter is 13 V
Sol: π°π =
π½ ππ
=
= π. ππ π¨
πΉ
π
Loop (bxcb)
(π × π°π ) + (π × π. ππ) − ππ = π → π°π = π π¨
ππ πππ
π π → π°π + π°π = π°π
π°π = π. ππ − π = π. ππ π¨
47) Four identical light bulbs K, L, M, and N are connected in
the electrical circuit shown in the accompanying figure. In
order of decreasing brightness (starting with the brightest),
the bulbs are:
βΆ K=L>M>N
β· K=L=M>N
βΈ N>K>L=M
ππ → ππ > ππ → πππ«ππ§ππ‘ > ππ = ππ
(π, π) π¬ππ«π’ππ¬
35
Mr. Hytham Ahmed
48) The potential difference between two ends of a conductor when a work of 1
Joule is done to transfer unit charge between them is.
a) Volt
b) potential difference
c) ohm
d) quantity of charge
49) If three copper wires x, y and z of lengths 2 m, 4 m and 1 m respectively
have the same cross-sectional area, then which of the following figures
represents the resistances
πΏ
Answer: (a) , R=πeπ΄
36
,
same πe an same A , so R α L
Ly>Lx>Lz
∴ Ry > Rx >Rz
Mr. Hytham Ahmed
50) in the opposite figure:
The equivalent resistance between X and Y equals
………
a) 2 Ω
b) 4 Ω
c) 6 Ω
d) 8 Ω
10×40
∴ Req= 10+40 =8β¦
51) Which of the following figures represent the relation between the
conductivity of the material of a conductor and its cross-section area? …………..
Answer: (d) The conductivity is a physical property of material
depends only on the type of material and temperature
37
Mr. Hytham Ahmed
52) Which of the following graphs represents the relation
between the value of the rheostat Rv and the consumed
electric power in it? ………….
Answer :(a)
Pw=
38
π½π
πΉπ½
V is constant (same battery) , so Pw α
π
πΉπ½
Mr. Hytham Ahmed
53) In the electric circuit shown in the figure:
What will happen to the brightness of the two bulbs (a
and b) at closing the switch k?
a) decreases, increases
b) turned off, increases
c) increases, decreases
d) decreases, decreases
When k is closed a is off , Req increases & I decreases ∴Vout Increases & V of
each of lamp a , b increases
54) The value of power delivered in the
circuit when
the key (K) is closed:
βΆ144 W
β·60 W
βΉ96 W
βΈ24 W
π
24
π΅
IT= π
+π
= 4+2 = 4A
39
, P= ππ΅ πΌπ‘ππ‘ππ = 24 × 4 = 96 πππ‘π‘
Mr. Hytham Ahmed
55) Which of the following graphs represents the relation between the potential
difference (V) across the ends of a conductor that carries DC current and the
time (t) ? ………..
Answer :(a) potential difference doesn’t depend on time
56) In the opposite figure, if the reading of
Ammeter equals 1A, so the electromotive force of
battery (VB) equals …………..
βΆ2V
β·10V
βΈ15V
βΉ 12V
I6β¦=1A ,
I3β¦= I6β¦ +IAmmeter =1+1=2A , I12β¦=
Itotal=2+1=3A
40
,
6×2
12
= 1π΄
12×6
VB=IT (Req+r)=3(12+6 + 1) =15 Volt
Mr. Hytham Ahmed
57) In the circuit shown, the unknown (X) equals
βΆ5
β·10
Vout=VB-Ir
, (10x1+ITx6)=24−IT x1 , ITx6+IT x1 =24-10
7IT=14
, IT=2A
βΈ6
, V//=IT Req //
βΉ30
10×π
10×π
, 10=2× 10+π
,
10+π
, R=10 β¦
=5
58) If the wire reshaped to decrease its diameter to half then its resistance
increase by……..
βΆfour times
β·constant
reshaped (Volume const), Vol=AxL
, R2=16R1 ,
41
R inc to 16 times
βΈ8 times
,
1
1
AπΌ(2)2 π 2
4
,
,
βΉ15 times
1
1
AπΌ 4πΏ
4
4Rα4L ,
R inc by=16R-R=15R
1
1
4
1
π΄
4
RπΌ
Mr. Hytham Ahmed
59) If the ratio between the current intensity passes through a
conductor and the potential difference across the conductor 0.5
A/V. The resistance of the conductor is ………. β¦
βΆ 3Ω
β΅
π°
π½
β·0.5Ω
π
π.π
πΉ
π
= =
, πΉ=
π
π.π
=π
βΈ5Ω
βΉ2Ω
∴R=2 β¦
60) Three resistors (60, 20, and 40) ohm are connected to an electric current
source. If the potential difference across each resistor is (30, 10, 20) volt
respectively, imagine how these resistors could be connected. Then the total
resistance of the electric circuit is………….
βΆ30Ω
R (β¦)
20
40
60
β·20Ω
V (V)
10
20
30
I (A)
0.5
0.5
0.5
βΈ60Ω
βΉ100Ω
40 Ω
20 Ω
60 Ω
(Req )series = 20 + 40 =60 β¦
π
60
60//60
Req = π = 2 = 30 β¦
42
Mr. Hytham Ahmed
61) Mathematical relation for ohm’s law closed circuit is…………….
βΆ
β· P=VI
V=IR
βΈV=VB-Ir
βΉ no one of them
62) If the reading of V1 = 80 volt and V2 = 48 volt , so the resistance
X equal ……..β¦
βΆ 3Ω
β· 5Ω
(12,8) series Req=12+8=20 Ω
πππ
,
8+X=12π
43
βΈ 8Ω
π½π
, I=
πΉ
=
ππ
ππ
βΉ 4Ω
= ππ¨
,
R(8+X)=
π½π
π°
=
ππ
π
=
X=4 π
Mr. Hytham Ahmed
63) If reading of ammeter when key is open =
2A,
and when closed reading becomes 2.4 A , then
internal resistance equal……
βΆ 1Ω
β· 2Ω
βΈ 0.5Ω
βΉ 1.5Ω
K is opened
,
π
π΅
I = 10+π
=2A
12// 12 = 6 ,
20 + 2r = VB
K is closed
,
π
π΅
I = 8+π
= 2.4 A
,
Req = 4 +6 = 10Ω
→(1)
12// 6 = 4 ,
19.2 + 2.4 r = VB
Req = 4 +4 = 8Ω
→ (2)
From (1&2)
20 + 2r =19.2 + 2.4 r
∴r=2Ω
64) In the opposite circuit, which of the following graphs represents
the relation between: the value of the resistance which is taken
from Rv and the reading of the ammeter?
1
Answer: (a) πΌ πΌ π
44
Mr. Hytham Ahmed
65) The reading of the voltmeter in the opposite figure equals ……….
a) 4 V
b) 6 V
c) 8 V
d) 12 V
π
3
→ π
ππ = π
+ 2 = 2 π
π
12
8
πΌ = π
π΅ = 3 = π
ππ
2
π
8
π = πΌπ
= × π
= 8 v
π
66) In the opposite figure:
If each resistor has a resistance R and
V2 reading is 33 V, so the reading of V1
is ………..
a) 9 V
b) 11 V
c) 3 V
d) 1V
→ π
(π
.π
.π
) = π
+ π
+ π
= 3π
π
(3π
//π
) =
3π
π
4π
3
= 4π
3
π
ππ = 4 π
+ π
+ π
=
π
33
12
π
π
πΌ = π
2 = 11 =
ππ
π = πΌπ
=
45
4
12
π
3
11
4
π
× 4π
= 9 v
--→ 9/3 = 3V
Mr. Hytham Ahmed
67) The opposite figure represents
three resistors (x, y, z) that are
connected in series. So, which of
the following figures represents the
ratios of the consumed electric power in each of them? …………
Answer: C π = πΌ 2 π
πΌ → ππππ π‘πππ‘
ππΌπ
46
Mr. Hytham Ahmed
68) Three equal resistors of 3 Ω each, are connected in series to a
cell of internal resistance 1 Ω. If these resistances are connected
in parallel and connected to the same cell, then the ratio of
respective currents through the electric circuits in the two cases is
………
a) 1 : 8
b) 1 : 7
c) 1 : 5
d) 1 : 3
SOL:-
ππ΅
π
→ πΌ1 = π
π πππππ +π
ππ΅
π
πΌ2 = π
πππ
πΌ1
πΌ2
=
ππ΅
10
ππ΅
2
π΅
= 9+1
=
π΅
= 1+1
=
+π
2
ππ΅
10
ππ΅
2
1
= 10 = 5
69) In the opposite figure the value of I is ………..
a) 2A
b) 4A
c) 6A
d) 12 A
6×12
→ π
(6//12) = 6+12 = 4πΊ
πΌ1 = πΌ2
πΌ = πΌ1 + πΌ2 = 2 + 2 = 4π΄
47
Mr. Hytham Ahmed
70) If the length of wire increase to double and its diameter
increases to double so its resistance…..
a) halved
b) doubled c) no change
d) increase 4 times
sol
π³π = ππ³π , π
π = ππ
π
πΉπ
πΉπ
πΉπ
πΉπ
πΉπ
πΉπ
=
ππ
=
π³π
π³π
×
ππ
π³π
π π
ππ
×
ππ³π
π
ππ
,
=π
×
π
ππ
π
ππ
π
= ×π
π
π
R2= R1
π
71) If (220 V - 100 W) is written on an electric fan and (220 V - 1000 W) is written
on an electric heater, then the resistance of the heater is ………
a) equal to the resistance of the fan
b) less than the resistance of the fan
c) greater than the resistance of the fan
d) indeterminable
→ π → ππππ π‘πππ‘
π=
ππ
πβ
π2
1
π
, π πΌπ
π
100
π
= π
β → 1000 = π
β
π
π
π
0.1 = π
β → π
β = 0.1 π
π
π
48
Mr. Hytham Ahmed
72) In the opposite figure:
If the electric current passing through the 2 Ω
resistor is 1 A, then the electric current
intensity passing in the 12 Ω resistor equals
………….
a) 0.5 A
c) 1.5 A
b) 1 A
d) 2 A
→ π
(2,4) = 2 + 4 = 6πΊ , πΌ6πΊ = 1π΄
6
1
π
(6//6) = 2 = 3πΊ , π
πΌ πΌ → πΌ3πΊ = 2π΄
π
(3,3) = 3 + 3 = 6πΊ
πΌ1
πΌ2
π
2
12
2
6
= π
2 → πΌ =
1
→ πΌ12πΊ = 1π΄
73) The opposite figure represents three resistors x, y, z
that are connected in parallel, so which of the
following figures represents the ratios of the current
intensity that passes in each of them?
Answer: A
49
1
π
πΌπΌ
Mr. Hytham Ahmed
74) Four identical electric bulbs A, B, C, D and E are connected
with a battery of negligible internal resistance as shown in
figure. If the potential difference between the terminals of the
bulb ( B ) is 2V, when key open; the electromotive force of the
battery when key closed equal …
βΆ 14 V
β· 12 V
βΈ8V
βΉ 15 V
VB = 4 + 2+ 4+4 = 14 V
75) Wire of resistance R if its length increase to double and diameter
decrease to half, so its resistance increased by…….
βΆ 4R
2R πΆπ L
β· 7R
4R πΆ
π
π
( )π π
π
π
βΈ8R
βΉ6R
R increase 8 times
R increase by = 8R- R = 7 R
50
Mr. Hytham Ahmed
76) The electric conductivity of a wire is ( ππ ) ,its volume is 4m3 and
its resistance is 4 β¦,then its area is……………
a) √
π
b)
ππ
π
√ππ
c) 3√ ππ
d) ππ √
π
ππ
77) Voltmeter reading when K1 & K2 open………… and when K1 open & K2
closed
(12 & 9.6 v,
zero & 8
K1 & K2 open
Req = zero
V = VB = 12 V
K1 open & K2 closed
π
12
Req = 4 + 4= 8
I = π
π΅+π = 8+2 = 1.2
ππ
51
9.6 & 8)
V = I R = 1.2 X 8 = 9.6 V
Mr. Hytham Ahmed
78) Quantity of heat produced after 10 minutes in a copper wire of
diameter 5mm and its length is 150 cm, an electric current pass in it
of intensity 4A and resistivity of copper 1.7x10-8 β¦m equal…………
(12.468 J,
R=
ππΏ
π΄
ππΏ
= π π2 =
1.7 π₯ 10−8 π₯ 1.5
π (2.5 π₯ 10−5 )2
3.11 J,
1.298 J )
= 1.298 x 10-3 β¦
E = I2 R t = 42 X 1.298 x 10-3 X 10 X 60 = 12.468 J
79) The opposite graph represents the relation between the
number of the electrons (N) that passes through a certain crosssection of a conductor in an electric circuit and the time (t), so
the graph that represents the relation between the current
intensity (I) that passes through the conductor and the time (t)
is…….
Answer :(a) certain cross-section (number of free electrons constant) so current
remain constant
52
Mr. Hytham Ahmed
80) In the shown electric circuit, the current
intensity passing through the 45 Ω resistors
equals …………
a) 2 A
b) 2.5 A
c) 4 A
d) 5 A
Req1 = 50 + 50 = 100
Req2 =100/2= 50
Req3=100 + 100= 200
Req4 =
Req5 =50 + 40 = 90
Req7 =45+45 = 90
I = V/R = 180 / 45 = 4 A
53
πππ π ππ
πππ+ππ
= 40
Req6 =90/2= 45
Req8 =90/2= 45
I45 = 4/2 = 2A
Mr. Hytham Ahmed
81) Wire of resistivity ρ and cross section area A is formed into an equivalent
triangular (XYZ) of Side b, the resistance between two vertices of the triangle X
and Y is………….
βΆ
π π
β·
π ππ
Req =
ππΉ .πΉ
ππΉ+πΉ
π π
π ππ
π
π ππ³
π
π π¨
= R=
βΈπ
=
π
ππ
βΉ
π ππ
π π
π ππ
π π¨
82) In the electric circuit shown in the figure, the equivalent
resistance becomes maximum value when closing switch ……….
βΆ1
βΈ3
β·2
βΉ4
1
R
3
2
54
4
Mr. Hytham Ahmed
R
83) A group of equal resistors connected in series, so their equivalent
resistance was 100 Ω and when they connected in parallel their
equivalent resistance become 4 Ω , then the value of one of them is
……….. Ω.
a. 20
b. 25
c. 100
d. 125
Req = nR = 100 Ω
Req=
πΉ
π
=4Ω
R = 4n
nR = n(4n) = 100
πππ
n2 =
π
= 25
4n2= 100
n= 5
R = 4n =4x 5 = 20
84) The electric resistivity of a wire is ( ππ ) ,its volume is 3m3 and its resistance
is 3 β¦,then its length is……………
a) √
πππ
L2 = R
55
ππ
π
b)
ππ
L = √πΉ
πππ
ππ
π
c)
√π π
= √π
π
ππ
=3√
π
ππ
π
ππ
√π
=
d) ππ √
π
ππ
π
√π π
Mr. Hytham Ahmed
85) Two electric bulbs of resistances R1 R2 respectively are
connected in parallel to an electric source, if R1 > R2 then
……….
a. the glow of bulb R1 is greater.
b. the glow of bulb R2 is greater.
c. the two bulbs the same glow.
d. no correct answer.
π=
π2
π
, so, since V is constant and π
1 > π
2 → π2 > π1 and the glow of bulb π
2 is
greater
86) Four lamps 6 Ω each, are connected in parallel. Then the
combination is connected to a 12 V battery with a negligible internal
resistance:
The total charge leaving the battery in 10 s is ………
a. 80 C
c. 40 C
b. 60 C
d. 20 C
π
6
= = 1.5ο
π 4
π
ππ΅
ππ΅ π₯π‘ 12π₯10
πΌ= =
→π=
=
= 80 πΆ
π‘ π
ππ
π
ππ
1.5
π
ππ =
56
Mr. Hytham Ahmed
87) In the opposite figure an equilateral triangle abc,
if I=6A, the current intensity passing in the side ac
equals …… Ampere.
a. 6
c. 3
b. 4
d. 2
Equilateral; Rab = Rbc = Rac, so we have 2R // R,
therefore Iab = 2 Iacb so, Iac = 2A
88) In the electric circuit shown in the figure, four
lamps are lighted, if the one pointed by the arrow is
blown up how many bulbs will be kept lighted?
a. 0
b. 1
c. 2
d. 3
57
Mr. Hytham Ahmed
89) In the electric circuit shown in the figure:
When putting the switch in position (1), a current
of 2 A passes in the ammeter, thus the value of
the resistance R is ………. Ω.
a. 30
b. 5
c. 7.5
d. 2.5
At 1: Req = 3/2 R, π° =
π½π©
πΉππ +π
ππ
→ π = π → πΉ = πο
πΉ
π
90) The equivalent resistance between the
points A, B when the switch K is opened
and when it is closed respectively is ………
a. 8 Ω,2 Ω
c. 8 Ω,4 Ω
b. 9 Ω,4 Ω
d. 36 Ω, 6 Ω
Opened: 12//6 + 6//12 → πΉππ = ππ
ππππ
ππ+π
= πο
Closed: (12//6 + 6//12) // 8 → πΉπ,ππ = ππ
→ πΉππ =
58
ππππ
ππ+π
= πο
π
= πο
π
Mr. Hytham Ahmed
91) wires made of cu. length and mass of 1st one is (10m, 0.1 kg) and for 2nd
(40m , 0.2kg) then ratio between R1 / R2 = ………….
a) 1/8
πΉπ
πΉπ
=
π³ππ
π³ππ
b) 1/4
=
ππ
ππ
→
πΉπ
πΉπ
c) 1/1
=
πππ
πππ
×
π.π
π.π
=
d) 1/16
π
π
92) What happens for the readings of devices shown on increasing
the value of variable resistor (R2)?
59
Reading of
ammeter (A)
Reading of
voltmeter
(V1)
Reading of
voltmeter (V)
A
Decreases
Decreases
Increases
B
Unchanged
Decreases
Unchanged
C
Decreases
Decreases
Decreases
D
Decreases
Increases
Increases
Mr. Hytham Ahmed
93) It equal numerically terminal voltage of battery when no current
flow in circuit
a) electromotive force
π½ = π½π© − π°π
b) internal resistance
πππππ π° = π ∴ π°π = π
c) resistance
∴ π½ = π½π©
94) When the current passing in an ohmic resistance is halved then
the consumed power:
a- remains the same
b- increases 2 times
c- increases 4 times
d-decrease to quarter
P = π°π πΉ
60
πΉ(ππππππππ) ∴ π πΆ π°π
Mr. Hytham Ahmed
95) what happen To total consumed power in parallel connection lamps if one
of them burned and its battery of neglected internal resistance.
a- remains the same
↑ πΉππ =
πΉ
↓π΅
b- increases
,↓π·=
c- decrease
π½π (πππππ )
↑πΉ
96) Which graph represents the relation between the
electromotive force of a battery and the total resistance
of the circuit?
V
Req
61
V
V
B
B
Req
Req
Mr. Hytham Ahmed
97) In the given electric circuit; on closing the key (K), the readings of both
ammeter and voltmeter …
Choice
Reading of
(A)
Increase
Decrease
Increase
Decrease
A
B
C
D
Reading of
(V)
Decrease
Increase
Increase
Decrease
Open:
Req = 6+3+2 = 11 β¦
π°=
π½π©
πΉππ +π
=
ππ
ππ+π
=ππ¨
(πππππππ ππππ
πππ)
π½πβ¦ = π°πΉ = π × π = ππ½
(π½ππππππππ ππππ
πππ)
Closed:
Req = (6//3) + 3 +2 = 7 β¦
π°=
π½π©
πΉππ +π
π°πβ¦ =
=
ππ
π+π
π°πΉπ
πΉπβ¦ +πΉπβ¦
=
= π. π π¨
π.π×π
π+π
= π. ππ¨
π½πβ¦ = π°πΉ = π. π × π = π. ππ½
62
(ππππππ ππππ
πππ)
(π½ππππππππ ππππ
πππ)
Mr. Hytham Ahmed
98) which lamp must cut off to obtain max
reading of voltmeter
(A – B – C)
π½ = π½π© − π°π
for V to be max π½ = π½π© , ππ π°π ππππ ππ ππππ
r has value so I must be zero
B must cutoff, so that it acts as an open key
99) In the circuit shown in the figure,
the value of R is ……. β¦
a. 4
c. 16
πΌ = 4π΄ ,
π
ππ =
ππ΅
πΌ
b. 10
d. 20
ππ΅ = 20 π
= 5β¦
10×π
π
ππ = (8 + 3//6)// π
= 10+π
= 5β¦
10 R = 5 (10+ R) = 50 + 5R
63
→ 5π
= 50 ∴ π
=
50
5
= 10β¦
Mr. Hytham Ahmed
100) In the show circuit, if the filament of the lamp
(P) is fired, the reading of the voltmeter….
a) increases.
b) decreases.
c) not change.
r=0
V
R
π = ππ΅ (ππ π = 0)
β΅ P // R
∴ π π π‘πππ πππ’ππ π‘π ππ΅
101) In the circuit shown, the value of resistance R which makes the
reading of the ammeter 2 A is equal to:
a) 2 Ω
b) 6 Ω
c) 8 Ω
d) 12 Ω
π° = ππ¨ ,
πΉππ =
πΉππ
64
π½π©
π½π© = ππ π½ ,
ππ
= πβ¦
ππΉ
=π+
π+πΉ
π°
=
π=π
π
→
π=
ππΉ
π+πΉ
→
ππΉ = ππ + ππΉ ∴ πΉ = π
Mr. Hytham Ahmed
P
102) Use the given figure to determine the ratio
between the readings of the two voltmeters
R
R
v1
v2
V1 , V2.
2,
(1,
π
πΌ π πππ ∴ π1 =
2
π
π
2
3)
=2
103) The opposite graph represents the relation
between the resistance (R) R and the length (β) of the
two copper wires (x) and (y), so the ratio between the
cross-sectional areas of the two wires
a)
c)
πΏ
π
= ππ΄
65
π
π
π
π
π
, π ππππ = π΄
b)
π
d)
√π
π
ππ±
ππ²
is ………
π
, π ππ π‘βπ π πππ
1
∴ π πππππΌ π΄
tan ππ¦
→ tan ππ₯ =
π΄π
π΄π
1
=3
Mr. Hytham Ahmed
R
104) In the opposite circuit when the switch K is closed, which of the
following choices represent the change in the reading of the
Ammeter and the Voltmeter?
Choice
Reading of
voltmeter
Reading of
Ammeter
r = 0.5β¦
a)
Increases
Increases
b)
Increases
Decreases
c)
Decreases
Increases
d)
No change
Increases
VB = 7V
A
3β¦
V
K
3β¦
Open:
π
ππ = 3 β¦
πΌ=π
ππ΅
ππ +π
7
= 3+0.5 = 2 π΄
π3β¦ = πΌπ
= 2 × 3 = 6π
(πππππ‘ππ πππππππ)
(ππππ‘πππ‘ππ πππππππ)
Closed:
3
Req = 2 β¦
πΌ=π
ππ΅
ππ +π
=3
2
7
+0.5
= 3.5 π΄
(ππππ‘ππ πππππππ)
π3β¦ = πΌπ
π = 3.5 ×
66
3
= 5.25π
2
(ππππ‘πππ‘ππ πππππππ)
Mr. Hytham Ahmed
105) In the opposite figure if the electric current passing
in resistor R1 is 2 A , then the equivalent resistance of the
circuit is …………
a) 3 Ω
b) 4 Ω
c) 6 Ω
d) 12 Ω
π°πΉπ = ππ¨ →
π° = ππ¨
→ πΉππ =
ππ
π
= πβ¦
106) In the opposite figure:
The equivalent resistance between A and B equals ……….
a) 3 Ω
b) 6 Ω
c) 9 Ω
d) 10 Ω
9
π
ππ = 3 + 3 = 6β¦
67
Mr. Hytham Ahmed
107) Which of the following graphs
represents the relation between the
readings of the ammeter A1 and A2 by
changing the rheostat value (R)?
(Notice that: I1 , I2 are drawn with the same scale)
108) In the opposite electric circuit: If the bulb lights
up with its full intensity, the equivalent resistance
between the two points x and y equals………
a) 0.45 Ω
b) 1 Ω
c) 5 Ω
d) 3 Ω
π
ππ’ππ =
→ πΌ=
π 2 1.52
=
= 5Ω,
π
0.45
4.5
= 1.5 π΄,
3
1.5
→ π
= 1.2 = 1.25Ω,
68
π3Ω = 6 − 1.5 = 4.5 π,
→ πΌππ’ππ =
1.5
= 0.3 π΄, πΌπ
= 1.5 − 0.3 = 1.2 π΄,
5
5π₯1.25
→ R eq = 5+1.25 = 1Ω
Mr. Hytham Ahmed
109) In the circuit shown: If V = 20 volt.
The ammeter reading is…………..A
a) 72
I=
π
π
=
b) 3
20
5
= 4π΄
c) 5
πππππππ ππ πππππ‘ππ =
4π₯30
10+30
d) 64
= 3π΄
110) in the following circuit: what's happen to brightness of lamps
when key is opened?
π) π π (π’π§ππ«πππ¬π), π π (ππππ«πππ¬π )
b) π π (ππππ«πππ¬π ), π π (π’π§ππ«πππ¬π)
c) π π (π’π§ππ«πππ¬π), π π (π’π§ππ«πππ¬π)
π
Key open (yb2a shelt branch) ↑ π
ππ = ↓ π
↑ π
ππ , ↑ ππ , ↑ ππ
∴↑ π2
(ππππ π‘πππ‘)ππ΅ =↑ ππ ↓ ππ
69
∴↓ π3
Mr. Hytham Ahmed
111) In the given electric circuit, what would happen
to the reading of both voltmeters (V1) and (V2) when
increasing the value of the variable resistance (R1)?
a) V2 (constant ),V1 (increase )
b) V1 (increase ) , V2 (decrease)
c) V1 (constant) , V2 (increase )
R1 increase , Req increase , I decreases
↑V1 = VB - ↓ Ir
↓V2 = ↓IR
112) The opposite graph show that the relation
between the current flowing in two wires (A)
and (B)& the voltage across each of them, If
the two wires of equal length and the crosssection area.
I
A
B
1) Which wire has greater resistance
a) B
V
b) A
c) A and B have equal resistance
Wire B since slope =
70
π
πΉ
Mr. Hytham Ahmed
113) If the two wires together connected in parallel with the electrical
source which one of them consume more power
I
A
B
V
a) B
b) A
Wire A , as π =
π2
π
π(ππππ π‘πππ‘)
c) A and B have equal power
1
∴ ππΌ π
114) From the electric circuit shown in the figure ,
calculate the potential difference between points a, b
if the value of R = 7.5 Ω
a) 2.5 V
b) 1.3 V
c) zero
d) no correct
answer
Vad = VB- IR
71
Vac= VB- IR
∴ Vad = Vac ∴ Vab= zero
Mr. Hytham Ahmed
115) in the shown electric circuit, Calculate:
a- The total resistance of the circuit.
a) 5β¦
b)2.5β¦
c) 1.7β¦
d) 12β¦
Req = 2.5+2.5=5β¦
b- The total current intensity that passes in the circuit.
a) 3A
b) 12A
c) 1.3A
d) 9A
I=
π½
πΉππ
=
ππ
π
= 3A
cThe potential difference across a & b.
a) 15V
b) 5V
c) 7.5 V
Vab = I × 2.5 = 3× 2.5 = 7.5 V
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d) 45V
Mr. Hytham Ahmed
116) In the electric circuit shown in the figure, the
electric current becomes minimum value when
closing switch …
a. 1
b. 2
c. 3
d. 4
When Switch 2 is closed, Req = Maximum = 2R
117) In the opposite figure: if the potential
difference between the two terminals of the
metallic ring is 4π
V, the resistance of the wire
that the metallic ring is made of equals…. β¦
a. π
c. ππ
b. 2π
d. ππ
R = V/I = 4π/2 = 2π β¦
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Mr. Hytham Ahmed
118) From the opposite circuit: The current
intensity passing through the battery 12 V is …
a) 1 A
c) 0.33 A
πΉππ =
b) 0.62 A
d) 0.67 A
ππππ
πππππ
+π+
= ππο
ππ + π
ππ + ππ
ππ = π½π©π³ − π½π©πΊ = ππ − π = ππ½
→π°=
π½π©
π
=
= π. ππ π¨
πΉππ + π ππ
119) In the opposite circuit the ratio
a.
c.
π
ππ
π
ππ
π½π
π½π
= ………
b.
π
d.
π
π
π
π°πΉ
π
π°πΉ
π½π = π½π© + π°ππ = π½π© +
π
π½π©
ππ½π© − π½π©
π½π©
π°=
=
=
πΉππ + ππ + ππ ππΉ + π πΉ + π πΉ ππΉ
π
π
π½π© ππ
π½π© π
→ π½π = ππ½π© −
=
,
π½π = π½π© +
=
π
π
π
π
π½π π π
π
→
= π
=
π½π π ππ ππ
ππ = π½π© − π°ππ = ππ½π© −
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Mr. Hytham Ahmed
120) The circuit in the figure contains three
identical light bulbs in series with a battery.
When switch (S) is closed, which of the following
occurs to the bulbs (1&2)?
Bulb(1)
Bulb(2)
βΆ Goes out
Gets
brighter
β· Gets brighter Goes out
βΈ Gets brighter Gets
dimmer
βΉ Gets dimmer Gets
brighter
Ans:
Series & parallel (Volt change as RV change)
-if RV ↑ , Req// ↑ , π// ↑ , Vseries ↓
-if RV ↓ , Req// ↓ , π// ↓ , Vseries ↑
-But parallel only
β«Ψ«Ψ§Ψ¨Ψͺβ¬Volt β« ΨͺΩΨΆΩβ¬RVβ«Ω
ΩΩ
Ψ§ ΨΊΩΨ±Ψͺβ¬
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Mr. Hytham Ahmed
121) A network of electrical components is connected across a battery of
negligible internal resistance as shown, the resistance of the variable resistor
is increased. What is the effect on the readings of the voltmeters (V1), (V2)
and (V3)?
βΆ
β·
βΈ
βΉ
V1
Decreased
Increased
Unchanged
Unchanged
V2
Increased
Decreased
Decreased
Increased
V3
Decreased
Increased
Increased
Decreased
bulbs stay the same brightness
122)What happens to light bulbs A, B in the circuit during the slider P
moving from X to Y? neglecting internal resistance
76
light bulb (A)
light bulb (B)
βΆ
Doesn’t change
Increase
β·
Increase
Increase
βΈ
Decrease
Doesn’t change
βΉ
Increase
Decrease
Mr. Hytham Ahmed
123) in the opposite figure, when connected three
voltmeters on the three resistances, the relation
between readings of voltmeters is:
124) A wire of mass m & length L is made of a material density π if it has
resistance R so, its electric conductivity can be calculated from the relation
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Mr. Hytham Ahmed
125)
126)
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Mr. Hytham Ahmed
127)
128)
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Mr. Hytham Ahmed
129) In the electric circuit shown in the figure, the ratio between the reading of
voltmeter V1 and the reading of voltmeter V2 is:
a) 4
b) 2
c) 1
d) 0.25
130) In the circuit shown in the figure, calculate the value of the current I2 and I3
e) Value of I2:
i) 6
ii) 5
iii) 4
f) Value of I3:
i) 2
ii) 1.6
iii) 4
80
Mr. Hytham Ahmed
131) The figure shown represents a branching in an
electric circuit. Find the current in the branch X and
the direction of the current in the branch Y
g) 3 A & downward
h) 3 A & upward
i) 2 A & upward
j) 2 A & downward
132)In the circuit diagram shown:
neglecting the internal resistance of the two
batteries (VB1 and VB2)
the reading of the ammeter (A):
k) 3 A
l) 5 A
m) 2 A
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Mr. Hytham Ahmed
133) In the electric circuit shown:
(neglecting the internal resistance of the batteries)
the reading of the ammeters (A1) and (A2) is
n) 0.5 & 0.25 A
o) 1 & 2 A
p) 0.75 & 0.4 A
136) The least electric current
passes through the circuit if a
battery is connected between:
q) Points (A & C)
r) Points (D & C)
s) Points (A & B)
t)
82
Points (A & D)
Mr. Hytham Ahmed
137) In the opposite figure the current intensities (I1)
&(I2) equals:
u) 1A,3A
v) 2A,2A
w) 3A,2A
x) 4A,1A
138) In the opposite figure:
The reading of ammeter is:
a. 4A
b. 3.5A
c. 3A
d. 2A
83
Mr. Hytham Ahmed
139) Consider the circuit below, knowing that all
resistances equal 10 β¦ & with neglecting the internal
resistance of the batteries, the potential at the point
(B) will be………
a. 3.33A
b. 2.22A
c. 5.45A
d. 1.15A
140) If the reading of the ammeter is zero, find R.
Answer:
Vbranch = Vbranch
VR = 6V
β΅ VB = VR+V2β¦
10 = 6+V2β¦
→ V2β¦ = 4V
I2β¦ = IT = V2β¦/R2β¦ = 4/2 = 2A
R = VR/IT = 6/2 = 3β¦
84
Mr. Hytham Ahmed
141) In the following circuit, find:
1) current in each resistance
2) power consumed in circuit
Ans
85
Mr. Hytham Ahmed
142) Find I1 ,I2,I3
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Mr. Hytham Ahmed
143) the opposite figure represents a
part of an electric circuit. BY using
Kirchhoff’s laws & taking into account
the current directions paths & the shown
Labels calculate
144) Find A1 & A2 with neglect internal
resistance of batteries.
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Mr. Hytham Ahmed
145)
146)
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Mr. Hytham Ahmed
147)
148) A uniform wire of resistance R is uniformly compressed along its length,
until its radius becomes n times the original radius. Now resistance of the wire
becomes:
a.
b.
c.
π
π4
π
π2
π
π
d. nR
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Mr. Hytham Ahmed
149) A metallic rod in the shape of a cylinder of cross-sectional area 2
cm2 & resistance 22.5 β¦ is pulled uniformly until its cross-sectional
area becomes 1.5 cm2 , so its resistance will be……….
e. 37 β¦
f. 40 β¦
g. 52 β¦
h. 56 β¦
150) The opposite figure shows 4
parts of wire that has same kind of
material but different cross-sectional
area as shown in the figure so which of the following graphs represents
the relation between current and distance from point A to point E ?
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Mr. Hytham Ahmed
151) In the shown figure, the are two same wire used in
an experiment. If the time which electrons pass
through a certain cross section area is the same. which
wire’s electrons have the higher current?
a. I1>I2
b. I1<I2
c. I1=I2
d. Can’t be determined (not enough givens)
152) A copper cable needs to carry a current of 200A with a power loss
of only 3 β¦/m. what is the required radius of the copper cable?
(resistivity copper is 1.7 × 10-8)
i. 0.21 cm
j. 0.85 cm
k. 3.2 cm
l. 4 cm
91
Mr. Hytham Ahmed
153) In the following circuit
m. A1=A2=A3
n. A1>A2>A3
o. A1<A2<A3
p. A1>A2=A3
154) For the following graph between (I) & (V) line
……..represents ohmic resistance.
q. A
r. B
s. C
92
Mr. Hytham Ahmed
155) In this circuit the reading
of ammeter = 0.4 A, so the
magnitude of resistance X
and Y is:
93
Mr. Hytham Ahmed
