(14%) Problem 7: An electronic toy is powered by three 1.58-V alkaline cells, ea
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(14%) Problem 7: An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.021 , and a 1.53-V carbon-zinc dry cell, whose internal resistance is 0.12 . All four cells are connected in series as part of a single-loop circuit. The resistance of the rest of the circuit is 5.45 - 33% Part (a) What current, I, in amperes, flows through the toy's circuit? Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining Sin tan coS asin atan acotan0 sinh( cosh)tanh) cotanh) 20 cotan % per attempt) detailed view acos 12 3 END oDegrees Radians DEL CLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% 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?Explanation / Answer
Given,
E1 = E2 = E3 = 1.58 V ; r1 = r2 = r3 = = 0.021 Ohm ; E4 = 1.53 V ; r4 = 0.12 Ohm ; R = 5.45 Ohm
a)The total emf of the circuit is:
E = E1 + E2 + E3 + E4
E = 1.58 + 1.58 + 1.58 + 1.53 = 6.27 V
All the resistances are in series so,
Req = 3 x 0.021 + 0.12 + 5.45 = 5.633 Ohm
So the current in the circuit will be:
I = E/R = 6.27/5.633 = 1.11 A
Hence, I = 1.11 A
b)P = I^2 R = 1.11^2 x 5.45 =
Hence, P = 6.72 W
c)R = R1 + R2 + R3 + R(batt)
P = I^2 R => I = sqrt (P/R)
I = sqrt (0.5/5.45) = 0.303 A
P = V^2/R => V = sqrt (P R)
V = sqrt (0.5 x 5.45) = 1.65 Volts
V = E - IE
1.65 = 6.27 - 0.303(R1 + R2 + R3 + R(batt))
R(batt) = 6.27 - 1.65/0.303 - 3 x 0.021
Hence, R(batt) = 15.18 Ohm
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