Physics 2402 Test #2 (8:30 Class) Fall 2004
1 The capacitors C1 (1.00x10 F), C2 (2.00x10-6 F) and C4 (4.00x10-6 F) are connected to a 20.0 V battery
as shown in the figure.
(a) Find the equivalent capacitance.
(b) Find the charge on capacitor C1.
(c) Find the potential difference across capacitor C4.
-6
C2
20V
C1
C4
Solution:
1 (a) The capacitors C1 and C2 are in parallel and can be replaced by C’ shown in Fig 1:
C’ = C1 +C2 = 3.00x10-6 (F).
The capacitors C4 and C’ are in series and can be replaced by C’’ shown in Fig. 2:
1
1
1
=
+
C ′′ C 4 C ′
⇒ C ′′ = 1.71× 10 −6 (F )
20V
20V
C’
C”
C4
Fig. 1
Fig.2
(b) The charge stored in C” is: Q” = C”∆V”= 1.71x10-6 (20) = 3.42x10-5 (C).
Since C” replaced C’ and C4 in series, these capacitors all have the same charge. The potential difference across C’ is:
∆V’ = Q’/C’ = 3.42x10-5/3.00x10-6 = 11.4 (V).
Since C’ replaced C1 and C2 in parallel each of these capacitors has the same potential difference as C’. Therefore,
∆V1 = 11.4 (V) and the charge on C1 is: Q1 = C1∆V1 = 1.00x10-6(11.4) = 1.14x10-5 (C)
(c) The potential difference across C4 is: ∆V4 = Q4/C4 = 3.42x10-5/4x10-6 = 8.55 (V)
2 The capacitors C1 (1.00x10-6 F) and C2 (2.00x10-6 F) are fully charged.
The currents in the 40.0 (V) battery and the 2.00 (Ω) resistor are given in
the figure and the arrows represent the direction of these currents. Find:
(a) The current passing through the 5 (Ω) resistor and mark its direction
with an arrow on the diagram.
(b) The Emf of the battery labeled “E”
(c) The resistance of the resistor labeled “R”
(d) Find the charge stored by the capacitor C2.
C1
7Ω
1Ω
in
= ∑ I out
⇒ 8 = I +5 ⇒
I = 3( A)
∑ ∆V = 0
⇒ = 35 − 3(5) − E-3(4 ) + 5(2 ) = 0 ⇒
5Ω
5A
h
loop
h
c
h
b
loop
⇒ + 40 − 8 R − 8(1) − 14 − 5(2 ) = 0 ⇒
R = 1.00(Ω )
40 V
E
4Ω
h
5Ω
I
h
35 V
h
h
14 V
a
A
2Ω
h
5A
going in the clockwise direction:
(iii) We apply Kirchoff’s Rule #2 around the outside loop going in a counterclockwise direction:
∑ ∆V = 0
8A
R
1Ω
E = 18.0(V )
2Ω
14 V
h
(ii) The high potential side of each resistor and battery are marked by
the symbol “h”. We apply Kirchoff’s Rule #2 to the inner loop “A”
4Ω
35 V
Solution:
2 (a) Since the capacitors are fully charged no charge is passing into the
capacitors. Therefore to analyze the circuit we ignore the presence of
the capacitors.
(i) We apply Kirchoff’s Rule #1 at the junction marked “a”. The current
“I” through the 5 ohm resistor must be in the direction shown:
∑I
40 V
E
12 Ω
C2
8A
R
(b) Since no current passes through the 12 ohm resistor the potential difference across it is zero. Therefore, the potential
difference across C2 is the same as the potential difference Vc-Vb where “c” and “b” are shown in the diagram. This
potential difference is found by adding up the potential differences as we move from “b” to “c” through the circuit:
Vc − Vb = +5(3) − 35 + 14 + 8(1) = 2.00(V )
The charge on C2 is: Q2 = C2∆V = 4.00x10-6 (C)
3(a) A charged capacitor C (4.00x10-6 F), a resistor R (3.00x10-6 Ω) and an open switch are connected
together as shown in Fig. a. The charge on the capacitor before the switch is closed is 1.20x10-4 (C). The
switch is now closed. How long will it take until the potential across the capacitor is 23.4 (V)?
C
switch
R
Fig a
6Ω
3Ω
5Ω
(b) This part of the problem is not related to part a.
(i) Find the equivalent resistance of the circuit shown in Fig. b.
(ii) If the current in the 3.00 ohm resistor is 2.00 (A), find the Emf, “E” of the battery.
E
Fig. b
Solution:
3 (a) The time constant “τ” of the circuit is: τ = RC = 12.0 (s).
The charge on the capacitor when its potential difference is 23.4 (V) is: q = CDV = 4x10-6(23.4) = 9.36x10-5 (C).
Let “t” be the required time and we use the theory for the discharge of a capacitor:
q (t ) = q0 e
−t
τ
⇒ 9.36 × 10 −5 = 1.2 × 10 − 4 e
−t
12
⇒ 0.78 = e
R
To find “t” we take natural logs:
−t
ln(.78) =
12
Part a
⇒ t = −12 ln (.78) = 2.98(s )
(b) (i) The 6 and 3 ohm resistors are in parallel and can be replaced by R’: 1/R’ = 1/6 + 1/3.
We find that R’ = 2.00 (Ω).
R’ and the 5 Ω resistors are in series and can be replaced by R": R” = R’ + 5 = 7.00 (Ω).
(ii) The potential difference across the 3 ohm resistor is: ∆V3 = I3(3) = 6.00 (V).
The potential across the 6 ohm resistor is the same since the 6 and 3 ohm resistors are in
parallel. The current in the 6 ohm resistor is: I6 = ∆V6/6 = 1.00 (A).
We let the current through the battery be ‘I”. By using Kirchoff’s First Rule at the junction
point “A” in Fig. b1 we have:
∑I
in
= ∑ I out
⇒ 1+ 2 = I
C
switch
−t
12
⇒
I = 3( A)
We apply Kirchoff’s Rule #2 to the loop of Fig. b3:
+E -3(7) = 0 and E = 21.0 (V).
I6
A
I
2A
6Ω
3Ω
Fig. b1
R’
5Ω
E
5 Ω R”
E
Fig. b2
E
Fig. b3
Physics 2402 Test #2 (11:30 Class) Fall 2004
1 The capacitors C1 (1.00x10 F), C2 (2.00x10-6 F) and C3 (3.00x10-6 F) are connected to a 20.0 (V) battery.
C2
(a) Find the equivalent capacitance of this circuit.
(b) Determine the charge on C2.
(c) Find the energy stored in C1.
-6
C1
20 V
C3
Solution:
1(a) The capacitors C2 and C3 are in parallel and can be replaced by the capacitor C’ shown in
Fig. 1:
C’ = C2 + C3 = 5.00x10-6 (F).
The capacitors C’ and C1 are in series and can be replaced by C” in Fig. 2:
1
1
1
= +
C ′′ C ′ C1
C’
C”
C1
⇒ C ′′ = 8.33 × 10 −7 (F )
20 V
Fig. 1
20 V
Fig. 2
(b) The charge on Q” is: Q” = C”∆V” = 8.33x10-7(20) = 1.67x10-5 (C).
Since C” replaced C’ and C1 in series, these capacitors all have the same charge. The potential difference across C’ is:
∆V’ = Q’/C’ = 1.67x10-5 /5x10-6 = 3.34 (V).
Since C’ replaced C2 and C3 in parallel each of these capacitors has the same potential difference of 3.14 (V). The charge
on C2 is:
Q2 = C2∆V2 = 2x10-6 (3.34) = 6.68x10-6 (C).
(c) The charge on C1 is the same as on C”: Q1 = 1.67x10-5 (C).
The potential difference is: ∆V1 = Q1 / C1 = 1.67x10-5 / 1x10-6 = 16.7 (V).
The energy stored is: U = 0.5Q1∆V1 = 1.39x10-4 (J).
2 (a) The capacitor C1 (1.00x10-6 F) is fully charged. The currents in the
49.0 (V) battery and the 20.0 (V) battery are given in the figure and the
arrows represent the direction of these currents.
Find:
(a) The current passing through the 2 (Ω) resistor and mark its direction
with an arrow on the diagram.
(b) The Emf of the battery labeled “E”
(c) The resistance of the resistor labeled “R”
(d) Find the potential difference across the capacitor C1.
12 V
E
C1
in
= ∑ I out
⇒ 7 = 3+ I
⇒
∑ ∆V = 0
⇒ + 20 − 3(5) − E + 4(2) + 4(1) = 0 ⇒ E = 17.0(V )
7A
R
49 V
12 V
5Ω
E
h
h
c
4Ω
⇒ + 49 − 7 R + 20 − 3(5) − 12 − 7(4 ) = 0 ⇒ R = 2.00(Ω )
3A
h
1Ω
h 7A
h
(c) We apply Kirchoff’s Rule #2 in the counterclockwise direction around the large outside loop:
h
20 V
I
25 V
49 V
∑ ∆V = 0
A
h
b
h
h
2Ω
loop
loop
3A
1Ω
25 V
I = 4( A)
(b) We apply Kirchoff’s Rule #2 in the counterclockwise direction
around the small loop “A”:
20 V
2Ω
8Ω
4Ω
Solution:
2 Since the capacitor is fully charged there is no current in that part of the
circuit so it has not been included in the figure.
(a) We apply Kirchoff’s First Rule at the junction labeled “a”. The
direction of the current in the 2 ohm resistor must be as shown by the
arrow:
∑I
5Ω
R
a
(d) Since no current exists in the capacitor part of the circuit there is no potential difference across the 8 ohm resistor. The
potential difference across the capacitor is the same as Vc – Vb.
Vc – Vb = +25- 4(2) + 17 – 12 = 22.0 (V).
3 (a) The capacitor C2 (2.00x10-6 F) is connected to a battery as shown in Fig. a1 until it is
fully charged. It is then disconnected from the battery (without changing the charge on the
capacitor) and is then connected to the capacitor C4 (4.00x10-6 F) as shown in Fig. a2. The
capacitor C4 was initially uncharged before it was connected to the other capacitor. After the
capacitors are connected the charge on C2 is 6.00x10-6 (C). Find the Emf, “E” of the battery.
(b) This part of the problem is not related to part a.
(i) Find the equivalent resistance of the circuit shown in Fig. b.
(ii) Find the power being dissipated as heat in the 4.00 ohm resistor.
E
C2
C2
C4
Fig. a1
Fig. a2
3Ω
20 V
6Ω
4Ω
Solution:
Fig. b
3 (a) In Fig. a2 the potential across C2 is: ∆V2 = Q2/C2 = 6x10-6/2x10-6 = 3.00 (V).
C2
E
C2
C4
The capacitors C2 and C4 must have the same potential difference. Therefore, the charge on
-6
C4 is: Q4 = C4∆V4 = 12.0x10 (C). Since the charge that was originally on C2 in Fig. a1 is
Fig. a1
Fig. a2
shared by C2 and C4 when connected together in Fig. a2, the total charge originally on C2 in
Fig. a1 is the sum of the charges on the capacitors in Fig. a2. The potential difference across C2 in Fig. a1 is: ∆V = Qtotal /
C2 = (6+12)x10-6/2x10-6 = 9.00 (V). Using Kirchoff’s second rule applied to the loop in Fig. a1 we have:
E – 9 = 0 and E = 9.00 (V).
(b) (i)The 4 and 6 ohm resistors are in parallel and can be replaced by R’ shown in Fig.
b1:
20 V
20 V
1 1 1
= +
R′ 4 6
⇒
3Ω
R′ = 2.40(Ω )
R’
R”
In Fig. b1 the 3 ohm resistor and R’ are in series and can be replaced by R” shown in
Fig. b2:
Fig. b1
Fig. b2
R” = 3 + R” = 5.40 (Ω).
(ii) The current passing through the battery is found using Kirchoff’s second rule applied to the circuit of Fig.
b2: +20 – I(5.4) = 0 and I = 3.70 (A)
The potential difference across R’ is: ∆V’ = I(R’) = 8.88 (V).
Since R’ replaced the 4 and 6 ohm resistors in parallel the potential difference across these resistors is also 8.88 (V).
The current in the 4 ohm resistor is: I4 = ∆V4/4 = 8.88/4 = 2.22 (A).
The power is: P4 = I42R = (2.222)4 = 19.7 (W)
Physics 2402 Test #2 (12:30 Class) Fall 2004
1 The capacitors C1 (1.00x10-6 F), C2 (2.00x10-6 F), C3 (3.00x10-6 F) and C4 (4.00x10-6 F) are
connected to a 20.0 V battery as shown in the circuit diagram.
(a) Find the equivalent capacitance.
(b) Find the potential difference across capacitor C4.
(c) Find the charge on capacitor C2.
C4
C3
20 V
C2
C1
Solution:
1 (a) The capacitors C1 and C2 are in parallel and can be replaced
by C’ shown in Fig. a:
C’ = C1 + C2 = 3.00x10-6 (F).
The capacitors C4 and C’ are in series and can be replaced by C”
shown in Fig. b:
C4
20 V
20 V
20 V
C3
C’
C3
C”
C”’
Fig. a
Fig. b
Fig. c
1
1
1
=
+
C ′′ C 4 C ′
⇒ C ′′ = 1.71 × 10 −6 (F )
The capacitors C3 and C” are in parallel and can be replaced by C”’ shown in Fig. c: C”’ = C3 + C” = 4.71x10-6 (F).
(b) The potential difference across the parallel capacitors C3 and C” is 20 (V). The charge on C” is:
Q” = C”∆V = (1.71x10-6)20 = 3.42x10-5 (C).
Since C” replaced C’ and C4 in series, each of these capacitors has the same charge as C”. The potential difference across
C4 is: ∆V4 = Q4/C4 = 3.42x10-5 / 4x10-6 = 8.55 (V).
(c) The charge on C’ is 3.42x10-5 (C) and the potential difference across it is: ∆V’ = Q’/C’ = 11.4 (V).
Since C’ replaced C1 and C2 in parallel these capaitors have the same potential difference as C’. The charge on C2 is:
Q2 = C2∆V2 = (2x10-6)11.2 = 2.24x10-5 (C).
2 The capacitor C (3.00x10-6 F) is fully charged. The currents in the
19.0 (V) battery and the 5.00 (Ω) resistor are given in the figure and the
arrows represent the direction of these currents.
Find:
(a) The current passing through the 4.00 (Ω) resistor and mark its
direction with an arrow on the diagram.
(b) The Emf of the battery labeled “E”
(c) The resistance of the resistor labeled “R”
(d) Find the potential difference across the capacitor C.
R
5Ω
E
19 V
in
= ∑ I out
⇒
∑
loop
10 V
3Ω
12 Ω
4Ω
29 V
R
h
h
h 5Ω
E
h
h
19 V
c
3Ω
h
10 V
A
h
b
5A
15 V
2A
h
E = 40.0(V )
C
5A
I = 5 + 2 = 7( A)
(ii) We apply Kirchoff’s second rule to loop “A” going in the clockwise
directio
∆V = 0 ⇒ + E − 2(5) − 15 − 10 − 7(4 ) + 29 − 2(3) = 0 n:
5V
2A
Solution:
2 Since the capacitor is fully charged, no current exists in that part of the
circuit and it is not shown in the figure.
(i) The current in the 4 ohm resistor must be in the direction of the
arrow. We use Kirchoff’s first rule at the junction “a”:
∑I
15 V
I
a
29 V
4Ω h
(iii) We apply Kirchoff’s second rule to the outer loop going in the
clockwise direction:
∑ ∆V = 0
⇒ + 19 − 5 R − 10 − 7(4 ) + 29 = 0
loop
R = 2.00(Ω )
(b) Since no current is passing through the 12 ohm resistor there is no potential difference across it. We do, however,
have to include the potential difference across the 5 volt battery in finding the potential difference across the capacitor.
The potential difference across the capacitor is the potential difference between points “c” and “b”:
Vc – Vb = -2(3) + 40 – 2(5) – 15 - 5 = 4.00 (V)
3 (a) The four resistors shown in the figure represent the resistances of the filaments
of four light bulbs in a battery powered emergency lighting system. The 3 ohm bulb
E
will be destroyed if the power dissipated in it exceeds 20.0 Watts.
(i) Find the equivalent resistance of this circuit.
(ii) Find the maximum possible Emf “E” of the battery if the 3 ohm bulb does not
burn out.
(b) This part of the problem is not related to part a. The switch “S” in Fig. b is initially open and
the capacitor is uncharged. Exactly 3.00 seconds after the switch is closed, the charge on the
capacitor is 6.20x10-5 (C). Find the resistance “R”.
1Ω
3Ω
6Ω
Fig. a
R
4Ω
4x10-6 F
S
20 V
Fig. b
Solution:
3 (a) (i) The 3 and 6 ohm resistances are in parallel and can be replaced
by R’ shown in Fig a2:
1 1 1
= +
R′ 3 6
⇒
R’
E
R′ = 2.00 (Ω )
4Ω E
R”
Fig. a1
Fig. a2
The three resistors in Fig. a2 are in series and can be replaced by R”
2.58
A
shown in Fig. a3:
1.29 A
I1
R” = 1 + 2 + 4 = 7.00 (Ω)
E
3Ω
6Ω
(ii) We let “I3” be the maximum current in the 3 ohm lamp which has a
1Ω
4Ω
power rating of 20 Watts.
P = I32R and I3 = (P/R)1/2 = (20/3)1/2 = 2.58 (A).
Fig. a3
The potential difference across the 3 ohm resistor is: ∆V3 = I3R = 7.74
(V).
Since The 3 and 6 ohm resistors are in parallel the potential difference across both resistors is 7.74 (V). The current in the
6 ohm resistor is:
I6 = ∆V6/R = 7.74/6 = 1.29 (A).
By applying Kirchoff’s rule #1 at the junction point between the 1, 3 and 6 ohm resistors in Fig. a1 we have:
∑I
in
= ∑ I out
⇒
I1 = I 3 + I 6 = 2.58 + 1.29 = 3.87( A)
We apply Kirchoff’s rule #2 to the circuit of Fig. a2. The current in this circuit is also I1.
E – I1R’ = 0 and E = 3.87(7) = 27.1 (V).
(b) When the capacitor is fully charged its charge is: Qf = CEbattery = 4x10-6(20) = 8.00x10-5 (C). The
charge three seconds after the switch was closed is:
−3

q (3) = Q f 1 − e RC

−3
−3


 ⇒ 6.2 × 10 −5 = 8 × 10 −5 1 − e RC




−3
.775 = 1 − e RC ⇒ .225 = e RC
R=
−3
= 5.03 × 10 5 (Ω )
C ln (.225)
⇒
−3
= ln (.225)
RC




R
S
Fig. b
4x10-6 F
20 V