Unit 2 Homework 2 Solutions
Internal Resistance
Total 35 marks
2000
24
a i) From the graph – extend the line to the y-axis E= 4.8V
ii) -r = gradient = (y2 – y1)/ (x2 – x1) (½) Points (0.4,4) (1.4,2)
= (2 – 4)/ (1.4 – 0.4) (½)
= -2
r = 2 Ω (1)
b i) Short Circuit Current = E/r
= 12/0.05
= 240 A
1
2
(½)
(½)
(1)
2
ii)
V1 = R1/(R1 + R2) X Vs
= 2.5/ (2.5 + 0.05) X 12
= 11.8V
V2/R
(½)
(½)
P=
(½)
= 11.82/2.5 (½)
= 55W
(1)
3
8
2002
10
B
24
a) 6 Joules of energy are transferred to each coulomb of charge by the
Source (1)
1
bi) I = 200mA
= 200 x 10-3
= 0.2 A
V = 6V
R=?
2
ii)
Emf = 1.5V
tpd = 1.3V
lost Volts = 0.2V
1
Total resistance R = V/I (½)
= 6/0.2
= 30 Ω (½)
R2 = 30 - 20 - 2= 8 Ω (1)
tpd = IR (½)
= 200 x 10-3 x 20 (½)
= 5.6 V
(1)
2
c) When switch S is closed, the two lamps are in parallel (½)
and the resistance decreases (½)
the share of the voltage decreases (½)
the voltmeter reading will be less (½)
2
8
2003
9
B
R = V/I =8/4 = 2 Ω
R = 2 – 0.2 = 1.8 Ω
1
24
bi)
RT = 0.5 + 1.5 = 2 Ω
I = V/R
30/2 (½)
= 15A (½)
1
P = I2R
= 152 x 0.5
= 112.5W
1
ii)
Internal resistance would share the voltage (½)
heating element voltage would be less (½)
Power output would be less (1)
OR
circuit resistance increases because of internal resistance r (½)
current decreases (½)
Power output would be less (1)
2
5
2004
24
a i)
lost volts = E – V = 9 - 7·8 = 1·2 (V) (½)
lost volts = Ir
1·2 = I x 2·0 (½)
I = 0·6 (A) (½)
R = V/I
= 7.8 /0.6 (½)
R = 13·0 Ω (1)
ii)
3
Current in internal resistor gives ‘lost volts’
(so external voltage reduced) (1)
1
OR External voltage or tpd = E – Ir
but as I increases and E and r constant then tpd decreases
OR Voltage divider explanation
eg voltage is divided across R and r so smaller reading on
meter
b i) (External) Resistors in parallel gives lower total resistance (½)
current increases (½)
so lost volts increases (½)
reading on voltmeter decreases (½) (0 if not in answer)
2
6
2005
25
a)
The energy given to each coulomb of charge passing through
the source/circuit
OR p.d. across the battery terminals when no current is drawn
/open circuit voltage/T.P.D + lost volts
1
b i)
A 6V
1
OR
B
ii)
E = V + Ir
(½)
6 =4.5 + 0.3 r (½)
R = 5.0 Ω
(1)
V = IR(½)
4·5 = 0·3 R (½)
R = 15 Ω
OR r = - gradient
suitable points from graph
2
1
c)
1=1 + 1
R 15 30
R = 10 (Ω) (1)
E = IR
6 = I (10 + 5) (½)
I = 0·4 (A) (½) (This is the reading on ammeter)
2
7
2006
9
B
Switch open RT = R1 + R2
=3+3=6Ω
Switch closed RT = R1 + Rp
=3+2=5Ω
(1/Rp = 1/3 + 1/6)
I=V/R = 12/2 = 6A
1
I=V/R = 12/5 = 4A
1