A battery with 5.00 v and no internal resistance supplies current to the circuit

Question

A battery with 5.00 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 1.07 mA. When the switch is closed in position a, the current in the battery is 1.27 mA. When the switch is closed in position b, the current in the battery is 2.10 mA. Find the following resistances. R2 R2 Rs (a) R1 4673 Your response differs from the correct answer by more than 100%. k (b) R2 (c) R3 kS2

Explanation / Answer

When the switch is opened,

Req = R1 + R2 + R3

Io = V/(R1 + R2 + R3)

(R1 + R2 + R3) = V/Io

at position A, the two R2 are in parallel

R2eq = R2/2

Req = R1 + R2/2 + R3

Ia = V/(R1 + R2/2 + R3)

R1 + R2/2 + R3 = V/Ia

when at b,

Req = R1 + R2

(R1 + R2) = V/Ib

we have three eqn solving that simultaneously, we get

R1 = V(1/Ib + 2/Ia - 2/Io) = 5(1/0.0021 + 2/0.00127 - 2/0.00107) = 909 Ohm

R2 = 2E(1/Io - 1/Ia) = 2 x 5 (1/0.00107 - 1/0.00127) = 1472 Ohm

R3 = V (1/Io - 1/Ib) = 5(1/0.00107 - 1/0.00127) = 736 ohm

Hence, R1 = 0.909 kOhm ; R2 = 1.472 kOhm ; R3 = 0.736 k Ohm

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