Question

(17%) Problem 4: An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.01852, and a 1.53-V carbon-zinc dry cell, whose internal resistance is 0.125 0. All four cells are connected in series as part of a single-loop circuit. The resistance of the rest of the circuit is 7.8 2. A 33% Part (a) What current, I. in amperes, flows through the toy's circuit? > A 33% Part (b) How much power. P, in watts, is supplied to the toy? P= Grade Summary Deductions 0% Potential 100% Late Work % 50% Late Potential 50% 7T ( 7 8 9 HOME E 4 5 6 sin cos tano cotan asino acos atan acotan sinh cosho tanho cotanh Degrees O Radians * 1 2 3 Submissions Attempts remaining: (0% per attempt) detailed view + 0 END NO BACKSPACE DAL CLEAR Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback. A 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. 72, failing, in ohms of the failed dry cell?

Answer 1

Total emp of all cells &=&, text&pt Eg {= 3X1.58+1.53= 6.27 V Total resistance of circuit Ref ritratra trg+R Rą=3(0.0185)+0.125+7-8 Rex = 7.98052 la) Current in ausrcuit (I) 6.27 I= E -0.7856A Rep 7.9805 I= 0.7856A b) fower supplied to toy P= I2R = (0.7856) ² (7.8) P= 4.834 c) New power to toy Pl_4.83-0.5 Pl= 4.3147 W so I'= pl I'= 0.7457 R so Rex = 2 = 8 so = 2 + (RG-Rey) = 0.572