(696) Problem 15: A 0.018-Q ammeter is placed in series with a 95- resistor in a
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Question
(696) Problem 15: A 0.018-Q ammeter is placed in series with a 95- resistor in a circuit. > 33% Part (a) Calculate the resistance, in ohms, of the combination. Grade Summary Deductions Potential 0% 100% sin0 cosO cotan0 asin acos0 atan0 acotan sinhO cosh0 tanh cotanh0 O Degrees Radians tan Submissions Attempts remaining: Z (490 per attempt) detailed view *1 2 3 END Submit Hint I give up! Hints: deduction per hint. Hints remaining: Feedback: 0%-deduction per feedback. - 33% Part (b) If the voltage is kept the same across the combination as it was through the 9.5- resistor alone, what is the percent decrease in current? - 33% Part (c) If the current is kept the same through the combination as it was through the 9.5- resistor alone, what is the percent increase in voltagei?Explanation / Answer
15)(a)
R3 = 0.018 + 9.5 = 9.518 Ohm
Hence, R3 = 9.518 Ohm
b)i1 = V/9.5
i2 = V/9.518
% decrease will be:
% = (i1 - i2)/i1 x 100
% = [(V/9.5 - V/9.518)/V/9.5] x 100 = 0.19 %
Hence, % = 0.19 %
c)i = V1/10
V1 = 10 i
i = V2/9.518
V2 = 9.518 i
V2/V1 = 9.518/10 = 0.952
% = (9.518 - 9.5)/9.518 = 0.189
Hence, % = 0.189
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