Cutnel, Physics, 10e Assignment> Open Assignment Chapter 20, Problem 062 HW # 4C

Question

Cutnel, Physics, 10e Assignment> Open Assignment Chapter 20, Problem 062 HW # 4Chap20Part2 033 A 77.3- resistor is connected in parallel with a 121.0· resistor. This parallel group is connected in series with a 26.2- resistor. The total combination is connected across 21.4-V battery. Find (a) the current and (b) the power dissipated in the 121.0- resistor 042 G0 050 320 20.Problem Chacter 20, Problem 04 G0 (a) Number Units Chacter 20, Problem (b) Number Units 20·Problem 00 Review Score Review Results by Study By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor

Explanation / Answer

for R1 and R2,

1/R = 1/77.3 + 1/121

R = 47.2 ohm

now R and R3 are in series.

Req = 47.2 + 26.2 = 73.4 ohm

current through battery, I = 21.4 / 73.4 = 0.292 A  

(A) through 121 ohm = 0.292 x 77.3 / (121 + 77.3)

= 0.114 A

(B) P = I^2 R =1.56 Watt

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