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julibyrd julibyrd
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11 years ago
The circuit diagram below shows two emf sources and a bulb connected in parallel. Also connected in the circuit is a resistor with resistance  R = 0.2 ohms . The resistance of the bulb is  R_b = 0.5 ohms, and each of the sources has internal resistance: r_1 = 0.025 ohms and  r_2 = 0.02 ohms .

http://session.masteringphysics.com/problemAsset/1060026/21/1060026A.jpg

If E1 = 13.0 V  and  E2= 5.0 V , calculate the current I_2 flowing in emf source E2.

*The direction of the current is Letter D in the link below*

http://session.masteringphysics.com/problemAsset/1060026/21/1060026B.jpg
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11 years ago
The potential differences in a CCW circuit of the upper loop must add up to zero.
The E1 may be regarded as being in series with a nearby resistor r1= 0.025 ohms,
and the E2 may be regarded as being in series with a nearby resistor r2 = 0.02 ohms.
Thus I2 r_2 - 5V + R I1 + I1 r_1 = 13 V.
Here the only unknowns are I1 and I2.
For the lower loop we can write
I2 r_2 - 5V = Rb Ib, where Ib is taken to flow clockwise.
Here the unknowns are I2 and Ib.
Finally, we known that Ib + I1 = I2,
so we have 3 equations in 3 unknowns.
Eliminating Ib from the "lower loop" equation, we get
I2 r_2 - 5V = Rb I2 - Rb I1
However, from the upper loop we have
I1 = (18V - I2 r_2) / (R+r_1).  Therefore,
I2 r_2 - 5V = Rb I2 + (I2 r_2 - 18 V) Rb / (R+r_1)
I2 (0.02 ohms) - 5V = I2 (0.2 ohms) + I2 (0.044444 ohms) - 40 V
35 V = I2 (0.2 ohms + 0.044444 ohms - 0.02 ohms)
= I2 (0.224444 ohms)
I2 = 156 amps

Do check my arith !
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