Class I Help Electric Circuits Begin Date: 10r7/2017 12:00:00 AM -- Due Date: 10
ID: 1778448 • Letter: C
Question
Class I Help Electric Circuits Begin Date: 10r7/2017 12:00:00 AM -- Due Date: 10/15/2017 11:59:00 PM End Date: 10/16/2017 12:00:00 AM (14%) Problem 7: An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.0195 , and a 1.53-V carbon-zinc dry cell, whose internal resistance is 0.115 . All four cells are connected in series as part of a single- loop circuit. The resistance of the rest of the circuit is 6625 . to view ble. 33% Part (a) what current, I, in amperes, flows through the toy's circuit? Grade Summary 8% 92 % Potential Attempts remaining: 8 cotanasinacos0 atan) acotan)sinh0 cosh0 tanh) cotanh0 oDegrees Radians % per attempt) detailed view 12 3 4% 4% DE CLEAR Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback. 33% Part (b) How much power, P, in watts, is supplied to the toy? 33% Part (c) The dry cell fails in such a way that it maintains its emf while its internal resistance increases greatly. That results in a reduced power of only 0.500 W supplied to the toy. What is the internal resistance, r2 failing, in ohms, of the failed dry cell? All content © 2017 Expert TA, LLCExplanation / Answer
A] current i = V/R = [3*1.58 +1.53]/(0.0195*3+0.115+6.625) = 0.92226 A
B] power = i^2R = 0.92226^2*6.625 = 5.635 W
c] i^2R = 0.500
i = sqrt(0.5/6.625) = 0.2747
total resistance = [3*1.58 +1.53]/0.2747 = 22.825
r2failing = 22.825 - 0.0195*3 -6.625 = 16.14 ohm answer