A battery with epsilon = 8.00 V and no internal resistance supplies current to t

Question

A battery with epsilon = 8.00 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 1.08 mA. When the switch is closed in position a, the current in the battery is 1.21 mA. When the switch is closed in position b, the current in the battery is 2.06 mA. Find the following resistances. (a) R_1 2291.82 Your response is off by a multiple of ten. k ohm (b) R_2 1591.67 Your response is off by a multiple of ten. k ohm (c) R_3 3523.91 Your response is off by a multiple of ten. k ohm

Explanation / Answer

when switch is open

R(total) = R1 + R2 + R3 = V/Io = 8/(1.08 x 10^-3) = 7407 ohm

when switch is in a

Rta = R1 + R2/2 + R3 = V/Ia = 8/(1.21 x 10^-3) = 6612 ohm

R2 = (7407 - 6612)*2 = 1584 ohm (1.584 k-ohm)

when switch is in b

Rtb = R1 + R2 = V/Ib = 8/(2.06 x 10^-3) = 3884 ohm

R1 = 3884 - 1584 = 2300 ohm (2.3 k-ohm)

R3 = (R1 + R2 + R3) - (R1 + R2) = 7407 - 3884 = 3523 ohm (3.523 k-ohm)

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