A battery with emf = 9.00 V and no internal resistance supplies current to the c

Question

A battery with emf = 9.00 V and no internal resistance supplies current to the circuit shown in the figure below. When the double-throw switch S is open as shown in the figure, the current in the battery is 1.00 mA. When the switch is closed in position a, the current in the battery is 1.18 mA. When the switch is closed in position b, the current in the battery is 1.86 mA. (a) Find the resistance R1. k? (b) Find the resistance R2. k? (c) Find the resistance R3. k?

Explanation / Answer

when open, current=E/(R1+R2+R3) =>1*10^-3=9/(R1+R2+R3) when a, current=E/(R1+R2/2+R3) =>1.18*10^-3=9/(R1+R2/2 +R3) when b, current=E/(R1+R2) =>1.86*10^-3=9/(R1+R2) solving, R1=4.161 kilo ohm R2=2.745 kilo ohm R3= 2.092 kilo ohm

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