Practice Problems (KEY)

Practice Problems (KEY)#

Solve the table for the empty cells based on the image below.

Variable	Circuit #1	Circuit #2	Circuit #3	Circuit #4
R1	500 \(\Omega\)	13.3 \(\Omega\)	2 k\(\Omega\)	100 \(\Omega\)
I1	40 mA	600 mA	6 mA	200 mA
V1	20 V	8 V	12 V	20 V
P1	800 mW	4.8 W	72 mW	4 W

For the circuit below, solve the table for the empty cells.

Variable	Circuit #1	Circuit #2	Circuit #3	Circuit #4
R1	50 k\(\Omega\)	53.3 M\(\Omega\)	2 M\(\Omega\)	400 \(\Omega\)
I1	100 \(\mu\)A	15 nA	500 \(\mu\)A	150 mA
V1	5 V	800 mV	1 kV	60 V
P1	500 \(\mu\)W	12 nW	500 mW	9 W

Which flashlight design is better if your main concern is power consumption?

Option 1	Option 2
R1 = 90 \(\Omega\)	R1 = 50 \(\Omega\)
V1 = 9 V	V1 = 6 V
P1 = 900 mW	P2 = 720 mW

Design 2 is better because 720 mW < 900 mW.

Two light bulbs, one rated at 50 W and the other rated at 100 W, are each connected to a 110 V source. Which bulb draws more current?

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

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

Bulb 2 draws more current because 909 mA is greater than 455 mA.

The lighting for a new UAS control console is modeled as a voltage source and a resistor, as shown below. For each of the following proposed options, find the unknown value. If the sole measure of merit is to minimize current, which of the six options is best?

Variable	Circuit #1	Circuit #2	Circuit #3	Circuit #4	Circuit #5	Circuit #6
R1	5.4 k\(\Omega\)	4.8 k\(\Omega\)	7.4 k\(\Omega\)	100 \(\Omega\)	667 \(\Omega\)	1.125 k\(\Omega\)
I1	2 mA	5 mA	1.08 mA	15 mA	8 mA
V1	10.8 V	24 V	8 V

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

The GPS receiver for an experimental cruise missile can be modeled as a single resistor connected to a voltage source. Given the six designs below, if the sole measure of merit is to minimize power consumption, which of the designs is best?

Variable	Circuit #1	Circuit #2	Circuit #3	Circuit #4	Circuit #5	Circuit #6
R1	100 \(\Omega\)	300 \(\Omega\)	8.2 k\(\Omega\)	4.8 k\(\Omega\)	1.5 k\(\Omega\)	1.2 k\(\Omega\)
I1	15 mA	8 mA	5 mA	12 mA	6 mA	10 mA
V1	1.5 V	2.4 V	41 V	57.6 V	9 V	12 V
P1	22.5 mW	19.2 mW	205 mW	691.2 mW	54 mW	120 mW

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

Given the following model of the cooling system of a laptop computer, clearly articulate how the current in the circuit would change if the resistance of the fan doubled from 1.5 k\(\Omega\) to 3.0 k\(\Omega\).

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.

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

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

Given the following model of an electric clock, clearly articulate how the power consumed by the clock would change if the voltage source decreased from 9.0 V to 4.5 V.

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.

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

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

To be effective as an illumination source, the thunderstorm lighting of an aircraft modeled as a single 10 \(\Omega\) resistor connected to a voltage source must consume a minimum of 250 W.

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

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

Not acceptable.

Alternate method:

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

Not acceptable.

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

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

A background light for a home stereo system is limited to 1 W to prevent damage to the stereo housing. The light is modeled as a single 100 \(\Omega\) resistor.

(a) Is a current of 9.2 mA acceptable?

\[ P = I^2 R = (9.2 mA)^2 \cdot 100 \Omega = 8.464 mW \]

Yes, this is acceptable.

Alternate method:

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

Yes, this is acceptable.

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

\[ I = \sqrt{\frac{P}{R}} = \sqrt{\frac{1 W}{100 \Omega}} = 100 mA \]