Homework 2 Questions

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University Of Connecticut *

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2000

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Electrical Engineering

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Jan 9, 2024

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1. Award: 10.00 points Consider a lead-acid battery having the following specifications, defined per U.S. standards: Voltage = 12 Volts RC = 160 minutes AH 20 = 80 Amp-hours CCA = 800 Amps How long can this battery power an electric heater that draws 10 Amps of current before it runs out? Please enter your answer in hours in the space provided, rounded to the tenths place. 7.3 ± 2.5%

Consider NiMH hobbyist batteries shown in the circuit in the given figures. Given: V 1 = V 2 = 14 V, R 1 = 0.17 Ω , R 0 = 2.57 Ω , and R 2 = 0.3 Ω NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. References Section Break Diﬃculty: Medium

2. Award: 5.00 points For Battery 1, in the first figure, find the load current I 0 and the power dissipated by the load. The load current I 0 is 5.11 ± 2% A. The power dissipated by the load is 67.09 ± 2% W. Explanation: The load current is: The power dissipated is: I 0 = 5.11 A P Load = 67.09 W References Numeric Response Difficulty: Medium = I 0 V 1 + R 1 R 0 = P Load I 2 0 R 0

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3. Award: 5.00 points If a second battery is connected in parallel with Battery 1 that has voltage V 2 = 14 V and R 2 = 0.3 Ω, determine the currents and the power dissipated. Current I 1 is 3.34 ± 2% A. Current I 2 is 1.89 ± 2% A. The load current I 0 of Battery 2 is 5.23 ± 2% A. The power dissipated by Battery 2 is 70.3 ± 2% W. Explanation: With another source in the circuit, we must find the new power dissipated by the load. To do so, we write KVL twice using mesh currents to obtain the values of currents I 1 and I 2 : The current load is: The power dissipated is: I 1 = 3.34 A I 2 = 1.89 A I 0 = 5.23 A P Load = 70.30 W References Numeric Response Difficulty: Medium + − − = 0 I 2 R 2 V 1 V 2 I 1 R 1 ( + ) + = I 1 I 2 R 0 I 2 R 2 V 2 = + I 0 I 1 I 2 = P Load I 2 0 R 0

4. Award: 10.00 points For the circuits in the figures below, determine the resistor values necessary to achieve the indicated voltages. Resistors are available in the 1/ 8, 1/ 4, 1/ 2, and 1 W ratings. The value of resistor R a is 10 ± 2% kΩ. The value of resistor R b is 12.2 ± 2% kΩ. The value of resistor R L is 101.25 ± 2% kΩ. Explanation: Resistor R a is calculated from: Resistor R b is calculated from: Resistor R L is calculated from: R a = 10 kΩ R b = 12.2 kΩ R L = 101.25 kΩ References Numeric Response Difficulty: Medium 20 = (50) R a + 15000 R a 2.25 = 5 × 10,000 10,000+ R b 2.83 = 110 × 2.7 × 10 3 2.7 × + 1 × + 10 3 10 3 R L

5. Award: 10.00 points The voltage divider network in the given figure is designed to provide v out = v s / 2. However, in practice, the resistors may not be perfectly matched, that is, their tolerances are such that the resistances are unlikely to be identical. Assume v s = 10 V and nominal resistance values R 1 = 205 k Ω and R 2 = 205 k Ω . NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. References Section Break Diﬃculty: Medium If the resistors have ±10 percent tolerance, find the expected range of possible output voltages. v out, min = 4.5 ± 2% V v out, max = 5.5 ± 2% V Explanation: 10 % worst case: low voltage R 2 = 184500 Ω, R 1 = 225500 Ω 10 % worst case: high voltage R 2 = 225500 Ω, R 1 = 184500 Ω v out, min = 4.50 V v out, max = 5.50 V References Numeric Response Difficulty: Medium = × 10 = 4.50 V v out, min 184500 184500 + 225500 = × 10 = 5.50 V v out, max 225500 184500 + 225500

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6. Award: 10.00 points Find the expected output voltage range for a tolerance of ±5 percent. v out, min = 4.75 ± 2% V v out, max = 5.25 ± 2% V Explanation: 5% worst case: low voltage R 2 = 194750 Ω, R 1 = 215250 Ω 5% worst case: high voltage R 2 = 215250 Ω, R 1 = 194750 Ω v out, min = 4.750 V v out, max = 5.250 V References Numeric Response Difficulty: Medium = × 10 = 4.750 V v out, min 194750 215250 + 194750 = × 10 = 5.250 V v out, max 215250 215250 + 194750

7. Award: 10.00 points Determine the equivalent resistance between terminals a and b in the given circuit. Assume R 1 = 8 Ω, R 2 = 7 Ω, R 3 = 13.25 Ω, R 4 = 7.75 Ω, R 5 = 4.5 Ω, R 6 = 5.5 Ω, R 7 = 4 Ω, and R 8 = 7 Ω. The equivalent resistance between terminals a and b is 11.6 ± 2% Ω. Explanation: The equivalent resistance can be determined as follows: The equivalent resistance between terminals a and b is 11.6 Ω. References Numeric Response Difficulty: Easy + = 4 Ω + 7 Ω = 11 Ω R 7 R 8 = 7 Ω 13.25 Ω 7.75 Ω = 2.88 Ω R 2 R 3 R 4 = 4.5 Ω 5.5 Ω = 2.48 Ω R 5 R 6 2.88 Ω + 2.48 Ω = 5.36 Ω 11 Ω 5.36 Ω = 3.6 Ω = + 3.6 Ω = 8 Ω + 3.6 Ω = 11.6 Ω R eq R 1

8. Award: 10.00 points Find the equivalent resistance seen by the voltage source and the current i in the given figure. Given R 1 = 1.6 Ω, R 2 = 5.5 Ω, R 3 = 34 Ω, R 4 = 150 Ω, R 5 = 11 Ω, R 6 = 5.5 Ω, R 7 = 5.5 Ω, and v S = 65 V. The equivalent resistance seen by the voltage source is 14.3764 ± 2% Ω. The current i is 0.3851 ± 2% A. Explanation: The equivalent resistance seen by the voltage source is Apply voltage division to find the voltage-drop across R 4 The current i through resistor R 4 is found using Ohm's law = + = + 1.6 Ω = 14.3764 Ω R eq + × + × × R 6 R 7 + R 6 R 7 R 3 R 5 + + × R 6 R 7 + R 6 R 7 R 3 R 5 R 2 R 4 + + + × × R 6 R 7 + R 6 R 7 R 3 R 5 + + × R 6 R 7 + R 6 R 7 R 3 R 5 R 2 R 4 R 1 +5.5 Ω ×150 Ω +34 Ω ×11 Ω 5.5 Ω×5.5 Ω 5.5 Ω+5.5 Ω +34 Ω +11 Ω 5.5 Ω×5.5 Ω 5.5 Ω+5.5 Ω +5.5 Ω +150 Ω +34 Ω ×11 Ω 5.5 Ω×5.5 Ω 5.5 Ω+5.5 Ω +34 Ω +11 Ω 5.5 Ω×5.5 Ω 5.5 Ω+5.5 Ω = × = 65 V × = 57.7659 V v R 4 v S − R eq R 1 R eq 14.3764 Ω−1.6 Ω 14.3764 Ω i = = = 0.3851 A v R 4 R 4 57.7659 V 150 Ω

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