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Kirchhoff's Junction Law

Three ammeter...

Question

Three ammeters A, B and C of resistances $R_{A}, R_{B}$ and $R_{C}$ respectively are joined as shown. 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

I_{A} = I_{B}

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

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

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

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Solution

The correct options are
A $I_{A} = I_{B}$
B $I_{A} R_{A} + I_{B} R_{B} = I_{C} R_{C}$
D $\frac{I_{B}}{I_{C}} = \frac{R_{C}}{R_{A} + R_{B}}$
According to the given data, ammeters A and B are connected in series hence will read the same current through them i.e.
$I_{A} = I_{B}$ ..........(1)
Now, applying KVL we get,
$I_{A} R_{A} + I_{B} R_{B} - I_{C} R_{C} = 0$
$I_{A} R_{A} + I_{B} R_{B} = I_{C} R_{C}$ ............(2)
Using equation (1) in (2) we get
$I_{B} R_{A} + I_{B} R_{B} = I_{C} R_{C}$
$I_{B} (R_{A} + R_{B}) = I_{C} R_{C}$
$\frac{I_{B}}{I_{C}} = \frac{R_{C}}{(R_{A} + R_{B})}$
$\frac{I_{A}}{I_{C}} = \frac{R_{C}}{(R_{A} + R_{B})}$ ........from(1)

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Three ammeters A, B and C of resistances RA,RB and RC respectively are joined as shown. When some potential difference is applied across the terminals T1 and T2, their readings are IA,IB and IC respectively

Three ammeters A, B and C of resistances $R_{A}, R_{B}$ and $R_{C}$ respectively are joined as shown. 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