A battery with emf = 7.00 V and no internal resistance supplies current to the c

Question

A battery with emf = 7.00 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 1.14 mA. When the switch is closed in position b, the current in the battery is 2.12 mA. Find the resistance R1. Your response differs from the correct answer by more than 10%. Double check your calculations. k? Find the resistance R2. Your response differs from the correct answer by more than 100%. k? Find the resistance R3.

Explanation / Answer

When the double-throw switch S is open as shown in the figure, the current in the battery is 1.00 mA

1*10^-3=7/[R1+R2+R3]

R1+R2+R3]=7000...(1)

When the switch is closed in position a, the current in the battery is 1.14 mA

1.14*10^-3=7/[R1+R3+0.5R2]

[R1+R3+0.5R2]=6140.351..(2)

When the switch is closed in position b, the current in the battery is 2.12 mA.

2.12*10^-3=7/{R1+R2}

R1+R2=3301.8868..(3)

SOLVING THREE (1),(2),(3) EQUATIONS WE GET

R1=1582.6 OHM=1.586 KOHM

R2=1719.3 OHM=1.72 KOHM

R3=3698.11 OHM=3.698 KOHM

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