HW6 Begin Date: 10/13/2017 12:00:00 PM-Due Date: 10/20/2017 11:59:00 PM End Date

Question

HW6 Begin Date: 10/13/2017 12:00:00 PM-Due Date: 10/20/2017 11:59:00 PM End Date: 10/20/2017 11:59:00 PM (7%) Problem 14: A 0.022-Q ammeter is placed in series with a 95- resistor in a circuit. 3396 Part (a) Calculate the resistance, in ohms, of the combination. 33% Part (b) If the voltage is kept the same across the combination as it was through the 95- 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 voltage? Grade Summary Deductions 100% Potential 0%

Explanation / Answer

Here ,

R1 = 0.022 Ohm

R2 = 9.5 Ohm

a) for the combination

resistance = R1 + r

resistance = 0.022 + 9.5

resistance = 9.522 Ohm

b)

let the voltage is 100 V

then current , Io = 100/9.50 = 10.523 A

when the ammeter is connected

I = 100/9.522 = 10.502 A

percentage decrease in current = (10.523 - 10.502) * 100/10.523

percentage decrease in current = 0.20 %

c)

for same current , I

initial voltage = 9.5I

final voltage = 9.522I

percentage increase in voltage = (9.522 - 9.5) * 100/9.50

percentage increase in voltage = 0.23 %

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