Practice Problems (KEY)

Problem 1

Variable	Circuit #1	Circuit #2	Circuit #3	Circuit #4
\(R_1\)	\(500\,\Omega\)	\(13.3\,\Omega\)	\(2\,\text{k}\Omega\)	\(100\,\Omega\)
\(I_1\)	\(40\,\text{mA}\)	\(600\,\text{mA}\)	\(6\,\text{mA}\)	\(200\,\text{mA}\)
\(V_1\)	\(20\,\text{V}\)	\(8\,\text{V}\)	\(12\,\text{V}\)	\(20\,\text{V}\)
\(P_1\)	\(800\,\text{mW}\)	\(4.8\,\text{W}\)	\(72\,\text{mW}\)	\(4\,\text{W}\)

Problem 2

Variable	Circuit #1	Circuit #2	Circuit #3	Circuit #4
\(R_1\)	\(50\,\text{k}\Omega\)	\(53.3\,\text{M}\Omega\)	\(2\,\text{M}\Omega\)	\(400\,\Omega\)
\(I_1\)	\(100\,\mu\text{A}\)	\(15\,\text{nA}\)	\(500\,\mu\text{A}\)	\(150\,\text{mA}\)
\(V_1\)	\(5\,\text{V}\)	\(800\,\text{mV}\)	\(1\,\text{kV}\)	\(60\,\text{V}\)
\(P_1\)	\(500\,\mu\text{W}\)	\(12\,\text{nW}\)	\(500\,\text{mW}\)	\(9\,\text{W}\)

Problem 3

Option 1	Option 2
R1 = 90 Ω	R1 = 50 Ω
V1 = 9 V	V1 = 6 V
P1 = 900 mW	P2 = 720 mW

Design 2 is better because 720 mW < 900 mW.

Problem 4

\[P_1 = 50\ \text{W} = V_1 I_1 \Rightarrow I_1 = \frac{50\ \text{W}}{110\ \text{V}} = 455\ \text{mA}\]

\[P_2 = 100\ \text{W} = V_2 I_2 \Rightarrow I_2 = \frac{100\ \text{W}}{110\ \text{V}} = 909\ \text{mA}\]

The 100 W bulb draws more current: 909 mA > 455 mA.

Problem 5

Variable	Circuit #1	Circuit #2	Circuit #3	Circuit #4	Circuit #5	Circuit #6
\(R_1\)	\(5.4\,\text{k}\Omega\)	\(4.8\,\text{k}\Omega\)	\(7.4\,\text{k}\Omega\)	\(4.8\,\text{k}\Omega\)	\(667\,\Omega\)	\(1.125\,\text{k}\Omega\)
\(I_1\)	\(2\,\text{mA}\)	\(5\,\text{mA}\)	\(1.08\,\text{mA}\)	1.04 mA	\(15\,\text{mA}\)	\(8\,\text{mA}\)
\(V_1\)	\(10.8\,\text{V}\)	\(24\,\text{V}\)	\(8\,\text{V}\)	\(5\,\text{V}\)	\(10\,\text{V}\)	\(9\,\text{V}\)

Circuit #4 is best because 1.04 mA is the smallest current.

Problem 6

Variable	Circuit #1	Circuit #2	Circuit #3	Circuit #4	Circuit #5	Circuit #6
\(R_1\)	\(100\,\Omega\)	\(300\,\Omega\)	\(8.2\,\text{k}\Omega\)	\(4.8\,\text{k}\Omega\)	\(1.5\,\text{k}\Omega\)	\(1.2\,\text{k}\Omega\)
\(I_1\)	\(15\,\text{mA}\)	\(8\,\text{mA}\)	\(5\,\text{mA}\)	\(12\,\text{mA}\)	\(6\,\text{mA}\)	\(10\,\text{mA}\)
\(V_1\)	\(1.5\,\text{V}\)	\(2.4\,\text{V}\)	\(41\,\text{V}\)	\(57.6\,\text{V}\)	\(9\,\text{V}\)	\(12\,\text{V}\)
\(P_1\)	\(22.5\,\text{mW}\)	19.2 mW	\(205\,\text{mW}\)	\(691.2\,\text{mW}\)	\(54\,\text{mW}\)	\(120\,\text{mW}\)

Circuit #2 is best because its power consumption of 19.2 mW is the smallest.

Problem 7

Resistance and current are inversely proportional, as shown by Ohm’s Law \(I = \frac{V}{R}\). Since voltage remains constant and resistance increases, current decreases.

\[I_{R_1} = \frac{V_S}{R_1} = \frac{12\ \text{V}}{1.5\ \text{k}\Omega} = 8\ \text{mA}\]

\[I_{R_2} = \frac{V_S}{R_2} = \frac{12\ \text{V}}{3.0\ \text{k}\Omega} = 4\ \text{mA}\]

If resistance doubles, current is halved.

Problem 8

Power is proportional to the square of voltage, as shown by \(P = \frac{V^2}{R}\). If voltage is halved, power is reduced by a factor of four.

\[P_1 = \frac{(9\ \text{V})^2}{3.2\ \text{k}\Omega} = 25.3\ \text{mW}\]

\[P_2 = \frac{(4.5\ \text{V})^2}{3.2\ \text{k}\Omega} = 6.33\ \text{mW}\]

Power decreases by a factor of 4 when voltage is halved.

Problem 9

(a) Would a 28 V battery be an acceptable voltage source?

\[P = \frac{V^2}{R} = \frac{(28\ \text{V})^2}{10\ \Omega} = 78.4\ \text{W} < 250\ \text{W}\]

Not acceptable.

Alternate method:

\[V = \sqrt{PR} = \sqrt{250\ \text{W} \cdot 10\ \Omega} = 50\ \text{V} > 28\ \text{V}\]

Not acceptable.

(b) What is the minimum allowable voltage for the voltage source?

\[V = \sqrt{PR} = \sqrt{250\ \text{W} \cdot 10\ \Omega} = \boxed{50\ \text{V}}\]

Problem 10

(a) Is a current of 9.2 mA acceptable?

\[P = I^2 R = (9.2\ \text{mA})^2 \cdot 100\ \Omega = 8.464\ \text{mW} < 1\ \text{W}\]

Yes, this is acceptable.

Alternate method:

\[I_{max} = \sqrt{\frac{P}{R}} = \sqrt{\frac{1\ \text{W}}{100\ \Omega}} = 100\ \text{mA} > 9.2\ \text{mA}\]

Yes, this is acceptable.

(b) What is the maximum allowable current in the circuit?

\[I_{max} = \sqrt{\frac{P}{R}} = \sqrt{\frac{1\ \text{W}}{100\ \Omega}} = \boxed{100\ \text{mA}}\]