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Potential Energy - External Viewpoint

Three ammeter...

Question

Three ammeters 1, 2 and 3 of resistances $R_{1}, R_{2}, R_{3}$ respectively are joined as shown. When some potential difference is applied across the terminals P and Q their readings are $I_{1}, I_{2}, I_{3}$ respectively. Then,

A

I_{1} = I_{2}

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B

I_{1} R_{1} + I_{2} R_{2} = I_{3} R_{3}

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C

\frac{I_{1}}{I_{2}} = \frac{R_{3}}{R_{1}}

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D

\frac{I_{2}}{I_{3}} = \frac{R_{3}}{R_{1} + R_{2}}

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Solution

The correct options are
A $I_{1} = I_{2}$
B $I_{1} R_{1} + I_{2} R_{2} = I_{3} R_{3}$
D $\frac{I_{2}}{I_{3}} = \frac{R_{3}}{R_{1} + R_{2}}$
As current in each capacitor in a branch is same so,
$I_{1} = I_{2}$
Both branches must have same potential difference as it is connected in parallel so,
$I_{1} R_{1} + I_{2} R_{2} = I_{3} R_{3}$
Now,
Let the potential difference from A to B is V.
Total current=I
Then,
$I_{3} = \frac{I (R_{3})}{R_{1} + R_{2} + R_{3}}$
$I_{1} = \frac{I (R_{1} + R_{2})}{R_{1} + R_{2} + R_{3}}$
So, $\frac{I_{2}}{I_{3}} = \frac{R_{3}}{R_{1} + R_{2}}$

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