# Solutions #2

```Physics 302/328 Homework (Part I)
Chapter 2, #1,7,9,10,11,14,25,26
2.1 Find the current I and the power supplied by the source in the
network shown below.
2
Chapter 2a.nb
Solution: I = V / R = 6 V / 20,000 W = .0003 A = 0.3 mA.
and P = V I = 6 V * 0.3 mA = 1.8 mW = 0.0018 W
2.7 A model for a standard two D-cell flashlight is shown in the figure
below. Find the power dissipated in the lamp.
Chapter 2a.nb
Solution: P = V 2 / R = H3.0 V L2 / 1 W = 9.0 W
2.9 Many years ago a string of Christmas tree lights was manufactured in
the form shown in digram (a) below. Today the lights are manufactured
as shown in diagram (b). Is there a good reason for this change?
3
4
Chapter 2a.nb
Solution: Since the builbs in (a) are in series, if one bulb burns out the current in all bulbs
becomes 0, and all the bulbs go out. The bulbs must be checked one by one to find the bad
bulb. In diagram (b), the bulbs are in parallel. If one bulb burns out, it is the only one to
stop glowing, and is easily replaced.
2.10 Find I1 in the circuit below.
Chapter 2a.nb
Solution: Looking at the bottom node, KCL gives us 12 mA - I1 - 4 mA - 2 mA = 0 =&gt; I1 = 6
mA
2.11 Find I1 and I2 in the circuit below.
Solution: Choosing a gaussian surface as shown, we have I1 -6 mA = 0 =&gt; I1 = 6 mA
Using KCL on the node that I1 flows into: 3 mA - I1 + I2 = 0 = 3 mA - 6 mA + I2 = 0=&gt; I2 = 3
mA
2.14 Find Ix in the circuit below.
5
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Solution: Using KCL on the top node, -2 mA + Ix - 2Ix + 4 mA = 0 =&gt; Ix = 2 mA
2.25 Find Vx in the circuit below.
Chapter 2a.nb
Chapter 2a.nb
7
Solution: Assume the current is flowing clockwise in the circuit.
KVL: 6V - 24 V + I (2 kW)+ I (5 kW) + I (2 kW) = 0 = -18 V + I (9 kW)
=&gt; I = 18 V / 9 000 W = 2 mA.
Vx = (2 mA) (5 kW) = 10 V
2.26 Find V1 in the circuit below.
V
Solution: KVL: – 25V + I (10 kW) + I (5 kW) + 4x = 0, with the current flowing clockwise in
the circuit.
Also, vx = – I (10 kW) , so
I H10 kW L
I (15 kW) –
= 25 V = I (12.5 kW), =&gt; I = 25V / 12.5 kW = 2.0 mA
4
Then vx = – I (10 kW) = ( - 2.0 mA) (10 kW) = - 20 V
V1 = 25 V + Vx = 25 V + (– 20 V) = 5V
```