Homework 1 Solutions PDF

Preview

Summary

Download Homework 1 Solutions PDF

Description

ECEN211 Homework 1 Solutions Published Jan. 23

HW1.1 (P1.31,P1.35,P.1.44 - slightly modified) (a)(5 points) Identify the nodes in the circuit of Figure P1.31. Keep in mind that all points connected by ideal conductors are considered to be a single node in electrical circuits. There are five nodes:

(b)(5 points) Identify elements that are in series. Elements B, D, and G are in series. (c)(5 points) Identify elements that are in parallel. Elements A and C are in parallel. (d)(5 points) What is the relationship between vA and vC ? vA = −vC . Because elements A and C are in parallel, the voltages are equal in magnitude, but because the reference polarities are opposite, the algebraic signs are opposite, too.

1

(e)(5 points) Given that vA = 8 V and vE = 4 V, determine the values of vC and vF . Using the KVL we can write the following equation going clockwise through elements A, E, and F : −vE − vC + vF = 0. Because vC = −vA we find vF = −4 V. HW1.2 (P1.57) A power of 100 W is delivered to a certain resistor when the applied voltage is 100 V. (a)(5 points) Find the resistance.

R=

1002 V12 = 100 Ω = 100 P1

(b)(5 points) Suppose that the voltage is reduced by 10 percent (to 90 V). By what percentage is the power reduced?

P2 =

V22 902 = 81 W = 100 R

for a 19% reduction in power. HW1.3 (P1.60)(10 points) The current through a 10-Ω resistor is given by i(t) = 10 exp(−2t) A. Determine the energy delivered to the resistor between t = 0 and t = ∞. The power is p(t) = i2 (t)R = 1000 exp(−4t). The energy is: Z ∞ Z ∞ p(t)dt = w= 1000 exp(−4t)dt = [−250 exp(−4t)]∞ 0 = 250 J 0

0

2...