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Inductors in Circuits

Two sources o...

Question

Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are $R_{1}$ and $R_{2}$ ( $R_{2}$ > $R_{1}$ ) . If the potential difference across the source having internal resistance $R_{2}$ is zero, then:

A

R = R_{2}

[ (

R_{1}

+

R_{2}

)/(

R_{1} - R_{2}

) ]

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B

R = R_{2} - R_{1}

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C

R = R_{1}

R_{2}

/(

R_{1}

+

R_{2}

)

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D

R = R_{1}

R_{2}

/(

R_{2} - R_{1}

)

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Solution

The correct option is A $R = R_{2} - R_{1}$
In the following figure:
$V_{b c} = E - i R_{2} = 0$ . ------------ (1)
The current $i$ flowing through the whole circuit is:
$i = \frac{2 E}{R + R_{1} + R_{2}}$ .
Using above equation in the first equation we get:
$E (1 - \frac{2 R_{2}}{R + R_{1} + R_{2}}) = 0$ .
Now, E is not equal to zero,
so the rest in the bracket must be zero i.e. $(1 - \frac{2 R_{2}}{R + R_{1} + R_{2}}) = 0$
or $R = R_{2} - R_{1}$ .

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