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

Question

A battery with emf 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 Io. When the switch is closed in position a, the current in the battery is Ia. When the switch is closed in position b, the current in the battery is Ib. (Use the following as necessary: script e for emf, Ia, Ib, and I0.)

http://www.webassign.net/serpse8/28-p-011.gif

(a) Find the resistance of R1.
R1 =

(b) Find the resistance of R2.
R2 =

(c) Find the resistance of R3.
R3 =

Explanation / Answer

when the switch is open, R1, R2 and R3 are in series, hence net current

I0  = e / (R1 + R2 + R3)

=> R1 + R2 + R3 = e / I0 ----------------- (1)

in position a for switch, R2 and R2 are in parallel and this combination is in series with R1 and R3

Ia   = e / (R1  + R2/2  + R3)

=> 2 * R1  + R2  + 2 * R3  = 2 * e / Ib   --------------- (2)

In position b, circuit has only R1 and R2 in series

Ib   = e / (R1  + R2)

R1 + R2   = e / Ib ---------------- (3)

Substituting this values in equation (1)

e/Ib   + R3   = e/I0

R3   = e * (1/I0  - 1/Ib)

then equation (2) gives

2 * R1  + R2  + 2 * {e * (1/I0  - 1/Ib)} = 2 * e / Ib

2 * R1  + R2   = 2 * e * {1 / Ib - (1/I0  - 1/Ib) }

2 * R1  + R2 = 2 * e * { 2 / Ib  - 1 / I0} ------------- (4)

solving equation (3) and (4)

R1  =   2 * e * { 2 / Ib  - 1 / I0} -   e / Ib

R1 = e * (3/Ib - 2/I0) ------------- (5)

Finally R2    = e/Ib   -   e * (3/Ib - 2/I0) from eq (3)

R2 = 2 * e * (1/I0  - 1/Ib)------------- (6)

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