(1496) Problem 7: A 0.0215-Q ammeter is placed in series with a 9.5- resistor in
ID: 1778697 • Letter: #
Question
(1496) Problem 7: A 0.0215-Q ammeter is placed in series with a 9.5- resistor in a circuit. @ 33% Part (a) Calculate the resistance, in ohms, of the combination. 33% Part (b) Ifthe voltage is kept the same across the combination as it was through the 9.5- resistor alone, what is the percent de in current? Grade Summary Deductions Potential Submissions Attempts remaining (iau per attempt cotan asinO acosO atanO acotan) sinh0 coshO tanh0cotanh0 derailed riew Degrees Radians Submit Hint I give up! Hints: 99' deduction per hint Hints remaining: Feedback: deduction per feedback. i33% Part (c) increase in voltage? If the current is kept the same through the combination as it was through the 10.00- resistor alone, what is the percentExplanation / Answer
a) RS = 0.0215 + 9.5 = 9.52 ohms
b) Is = V / R = V / 9.52
% decrease = [IR - Is] / IR * 100
= [V/9.5 - V/9.52] / [V/9.5] * 100
= 0.02 / 9.52 * 100
= 0.21 %
c) VR = I (9.5)
Vs = I (9.52)
% increase = I (9.52) - I (9.5) / I (9.5) * 100
= 0.02 / 9.5 * 100
= 0.211 %
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