1. Suppose a dc circuit contains three resistances

*R*1 = 22 ohms,*R*2 = 47 ohms, and*R*3 = 68 ohms across a battery that supplies a voltage of*E*= 3.0 V. Find the power dissipated by each resistance.*P*1 = 9.0/22 =

0.4091 W,

*P*2 = 9.0/47 = 0.1915 W, and*P*3 = 9.0/68 = 0.1324 W. These should be rounded off to*P*1 = 0.41 W,

*P*2 = 0.19 W, and

*P*3 = 0.13 W

2. Suppose we have a series circuit with a supply of 150 V
and three resistances:

*R*1 = 330 ohms,*R*2 = 680 ohms , and*R*3 = 910 ohms. What is the power dissipated by*R*2?
Round this off to three significant digits,
because that’s all we have in the data,

to obtain 4.15 W.

3. Suppose three resistors are in parallel across a battery
that supplies

*E*= 12 V. The resistances are*R*1 = 22 ohms,*R*2 = 47 ohms, and*R*3 = 68 ohms. These resistances carry currents*I*1,*I*2, and*I*3, respectively. What is the current,*I*3, through*R*3?*I*3 =

*E/R*3 = 12/68 = 0.18 A.

4. In the scenario shown by Fig. 1, what is the voltage
across the combination of

*R*3 and*R*4?
Ans. 5 V

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