MASSACHUSETTS INSTITUTE OF TECHNOLOGY
ESG Physics
8.02 with Kai
Spring 2003
Problem Set 6 Solution
Problem 1: 28.6
(a) Find the equivalent resistance between points a and b in Figure P28.6.
(b) Calculate the current in each resistor if a potential difference of 34.0 V is applied
between points a and b.
Solution:
(a)
Rp =
1
= 4.12Ω
1
1
+
7.00 10.0
Rs = R1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω
(1.1)
(1.2)
(b) For the 4.00 Ω and 9.00 Ω resistors, since ∆V = IR , we have
I=
∆V 34.0
=
= 1.99 A
R
17.1
(1.3)
For the 7.00 Ω resistor, we have
8.18
= 1.17 A
7.00
(1.4)
8.18
= 0.818 A
10.0
(1.5)
I=
For the 10.0 Ω resistor, we have
I=
Problem 2: 28.16.
Two resistors connected in series have an equivalent resistance of 690 Ω . When they are
connected in parallel, their equivalent resistance is 150 Ω . Find the resistance of each
resistor.
Solution:
Denoting the two resistors as x and y,
x + y = 690
(2.1)
1
1 1
= +
150 x y
(2.2)
and
Substituting equation (2.1) to (2.2), we have
( 690 − x ) + x
1
1
1
= +
=
x ( 690 − x )
150 x 690 − x
(2.3)
x 2 − 690 x + 103,500 = 0
(2.4)
which gives
By solving this quadratic equation, we have
x = 470 Ω and y = 220 Ω
(2.5)
Problem 3: 28.17
In Figures 28.4 and 28.5, let R1 = 11.0 Ω , let R2 = 22.0 Ω , and let the battery have a
terminal voltage of 33.0 V.
(a) in the parallel circuit shown in Figure 28.5, which resistor uses more power?
(b) Verify that the sum of the power ( I 2 R ) used by each resistor equals the power
supplied by the battery ( I ∆V ) .
(c) In the series circuit, which resistor uses more power?
(d) Verify that the sum of the power ( I 2 R ) used by each resistor equals the power
supplied by the battery ( P = I ∆V ) .
(e) Which circuit configuration uses more power?
Solution:
(a)
∆V = IR
(3.1)
33.0 V=I1 (11.0 Ω ) and 33.0 V = I 2 ( 22.0 Ω )
(3.2)
I1 = 3.00 A and I 2 = 1.50 A
(3.3)
Therefore
Therefore,
Since P = I 2 R , we have
P1 = ( 3.00 A ) (11.0 Ω ) and P2 = (1.50 A ) ( 22.0 Ω )
(3.4)
P1 = 99.0 W and P2 = 49.5 W
(3.5)
2
2
which gives
Therefore, the 11.0- Ω resistor uses more power.
(b) Since P1 + P2 = 148 W , we have
P = I ( ∆V ) = ( 4.50 )( 33.0 ) = 148 W
(3.6)
Rs = R1 + R2 = 11.0 + 22.0 = 33.0 Ω
(3.7)
(c)
Since ∆V = IR ,
I=
V 33.0 Ω
=
= 1.00 A
P 33.0 V
(3.8)
Since P = I 2 R , we have
P1 = (1.00 A ) (11.0 Ω ) and P2 = (1.00 A ) ( 22.0 Ω )
2
2
(3.9)
and you will probably get
P1 = 110 W and P2 = 22.0 W
(3.10)
Therefore the 22.0- Ω resistor uses more power.
(d)
P1 + P2 = I 2 ( R1 + R2 ) = (1.00 A ) ( 33.0 Ω ) = 33.0 W
(3.11)
P = I ( ∆V ) = (1.00 A )( 33.0 V ) = 33.0 W
(3.12)
2
(e) The parallel configuration uses more power.
Problem 4: 28.18
The ammeter shown in Figure P28.18 reads 2.00 A. Find I1 , I 2 and ε .
Solution:
Applying Kirchhoff’s Rule in the upper loop, we get
15.0 − ( 7.00 ) I1 − ( 2.00 )( 5.00 ) = 0
(4.1)
I1 = 0.714 A
(4.2)
I1 + I 2 = 2 A
(4.3)
I 2 = 1.29 A
(4.4)
which gives
Since
therefore
Applying Kirchhoff’s Rule in the bottom loop, we get
+ε − 2.00 (1.29 ) − ( 5.00 )( 2.00 ) = 0
(4.5)
ε = 12.6 V
(4.6)
which gives
Problem 5: 28.23
If R = 1.00 kΩ and ε = 250 V in Figure P28.23, determine the direction and magnitude
of the current in the horizontal wire between a and e.
Solution:
Label the currents in the branches as shown in the first
figure. Reduce the circuit by combining the two
parallel resistors as shown in the second figure.
Apply Kirchhoff’s looop rule to both loops in Figure
(b) to obtain
(5.1)
( 2.71R ) I1 + (1.71R ) I 2 = 250
(1.71R ) I1 + ( 3.71R ) I 2 = 500
(5.2)
With R = 1000 Ω , simultaneous solution of these
equations yields
I1 = 10.0 mA and I 2 = 130.0 mA
(5.3)
Vc − Va = ( I1 + I 2 )(1.71R ) = 240 V
(5.4)
From Figure (b),
Therefore, from Figure (a),
I4 =
Vc − Va
240 V
=
4R
4000 Ω
(5.5)
I 4 = 60.0 mA
(5.6)
which gives
Finally, applying Kirchhoff’s point rule at point a in Figure (a) gives
I = I 4 − I1 = 60.0 − 10.0 = 50.0 mA
(5.7)
I = 50.0 mA flowing from point a to point e
(5.8)
or
Problem 6: 28.34
A 4.00- MΩ resistor and 3.00- µ F capacitor are connected in series with a 12.0-V power
supply.
(a) What is the time constant for the circuit?
(b) Express the current in the circuit and the charge on the capacitor as functions of
time.
Solution:
(a)
τ = RC = ( 4.00 ×10−6 Ω )( 3.00 × 10−6 F )
(6.1)
τ = 12.0s
(6.2)
which gives
(b) sdfldskfj
I=
ε
R
e
−
t
τ
=
t
−
12.0
12.0
e
4.00 × 10−6
t
t
− 
−



q = Cε 1 − e τ  = 3.00 × 10−6 (12.0 )  1 − e 12.0 




(6.3)
(6.4)
After simplifying, we have
t
t
−
−


12.0
12.0
q = 36.0 µ C 1 − e
 and I = 3.00µ Ae


(6.5)
Problem 7: 28.69
(a) Using symmetry arguments, show that the current through any resistor in the
I
I
configuration of Figure P28.69 is either
of l All resistors have the same
3
6
resistance r.
5
(b) Show that the equivalent resistance between points a and b is r .
6
Solution:
(a)
r
r
i/6
r
r
r
a
r
r
b
r
i/6
i/3
a
i/3
3 junctions at
the same potential
(b)
r
i/3
i/3
b
i/6
r
r
i/6
i/6
r
another set of
3 junctions at
the same potential
i/3
i/6
i/3