Question

Engr 260--Circuits& Devices-Final May 21, 2019 Name: Instructions: You are allowed to use a calculator, the sheet containing the Second Order Circuits tables, and one 8.5" by 11" cheat sheet. You may not use a cell phone. 1) (2 pts) How much energy is delivered to the headlight below if it is left on overnight for 8 hours? Express your answer in kW-hours. 2) (6 pts) For the circuit below: a) what is the power associated with the current source? Is it absorbing or releasing (supplying) power? b) what is the power associated with the voltage source? Is it absorbing or releasing (supplying) power? c) what is the power dissipated by the resister? 2A 75V 10 2

Answer 1

1. Voltage V= 12V, Current I=5A

Power = Voltage * Current =12*5=60Watt

Energy = Power * Time=60*8=480WattHour

Voltage V = 12V, Current I = 5A Power = Voltage * Current = 12 * 5 = 60 Watt Energy = Power * Time = 60 * 8 = 480 Watt Hour = 480 * 10^-3 kWH = 0.480 kWH a. Current through resistor = 2A Let voltage across 2A current source be V_2A, Voltage across 10 Ohm resistor be V_10 Ohm By KVL - 75V - V_2A + V_10 Ohm = 0 Since V_10 Ohm = 2 * 10 = 20V V_2A = 20 - 75 = -55V Since V_2A is negative, Current source is absorbing power Power associated with current source = 55 * 2 = 110 Watt b. 75V voltage source is releasing power because current through voltage source is coming from its positive terminal. Power associated with 75V voltage source = 75 * 2 = 150 Watt c. Power dissipated in resistor = Power released by voltage source + Power absorb by current source

2.

a. Current through resistor=2A

Let voltage across 2A current source be $V_{2A}$ , Voltage across 10Ohm resistor be

By KVL $-75V-V_{2A}+V_{10\Omega}=0$

Since

Since $V_{2A}$ is negative, Current source is absorbing power

Power associated with current source=

b. 75V voltage source is releasing power because current through voltage source is coming from its positive terminal.

Power associated with 75V voltage source =

c.Power dissipated in resistor= Power released by voltage source + Power absorb by current source=

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