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Three ammeters A, B and C of resistances $R_{A}$ , $R_{B}$ and $R_{C}$ respectively are joined as shown in the figure. When some potential difference is applied across the terminals $T_{1}$ and $T_{2}$ their readings are $I_{A}, I_{B}$ and $I_{C}$ respectively. Then,

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$I_{A} R_{A} + I_{A} R_{B} = I_{C} R_{C}$

$\frac{I_{A}}{I_{C}} = \frac{R_{C}}{R_{A}}$

$\frac{I_{B}}{I_{C}} = \frac{R_{C}}{R_{A} + R_{B}}$

$I_{A} = l_{B}$

answer is A, B, D.

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Detailed Solution

(a) In series current is same.
$∴ I_{A} = l_{B}$
(b) $V_{A} + V_{B} = V_{C}$ ⇒ $∴ I_{A} R_{A} + I_{A} R_{B} = I_{C} R_{C}$
(d) In parallel current distributes in inverse ratio of resistance.

$\frac{I_{B}}{I_{C}} = \frac{I_{A}}{I_{C}} = \frac{R_{C}}{R_{A} + R_{B}}$

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