A battery with and no internal resistance supplies current to the circuit shown

Question

A battery with

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 1.05 mA. When the switch is closed in position a, the current in the battery is 1.28 mA. When the switch is closed in position b, the current in the battery is 2.05 mA. Find the following resistances.

A battery with emf = 4.00 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 1.05 mA. When the switch is closed in position a, the current in the battery is 1.28 mA. When the switch is closed in position b, the current in the battery is 2.05 mA. Find the following resistances.

Explanation / Answer

when S is open:

current=V/(R1+R2+R3)

R1+R2+R3=3.809 kilo ohms...(1)

when S is in position a,

two R2's are in parallel.

current=V/(R1+0.5*R2+R3)

R1+0.5*R2+R3=3.125 kilo ohms...(2)

when in b,

R3 is short circuited.

so current=4/(R1+R2)

hence R1+R2=1.9512 kilo ohms..(3)

solving the three equations

we get

R1=0.5832 kilo ohms

R2=1.368 kilo ohms

R3= 1.8578 kilo ohms

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