Question

The reading of the (ideal) ammeter, in the circuit shown here, equals:

(i) $I$ when key ${{\text{K}}_1}$ is closed but key ${{\text{K}}_2}$ is open.

(ii) $\dfrac{I}{2}$ when both keys ${{\text{K}}_1}$ and ${{\text{K}}_2}$ are closed.

Find the expression for the resistance of X in terms of the resistance of R and S.

Answer 1

We are asked to find the resistance of X in terms of the resistance of R and S. There are two conditions given, take each condition one by one and form equations for these two conditions using the concept of current and resistance. And use these equations to find the required answer.

Complete step by step answer:

Given a circuit, We are asked to find the resistance of X in terms of the resistance of R and S.

(i) When the key ${{\text{K}}_1}$ is closed but key ${{\text{K}}_2}$ is open.In this case current will flow through the resistances R and X. The resistances are in series, so equivalent resistance will be,

${R_{eq}} = R + X$

Current through a circuit is given by,

${\text{current}} = \dfrac{{{\text{voltage}}}}{{{\text{resistance}}}}$ (i)

Here, current is $I$, voltage is E and resistance is ${R_{eq}}$. Putting these values in the above equation we get,

$I = \dfrac{E}{{{R_{eq}}}}$

Putting the value of ${R_{eq}}$ in the above equation we get,

$\therefore I = \dfrac{E}{{R + X}}$ (ii)

(ii) When both keys ${{\text{K}}_1}$ and ${{\text{K}}_2}$ are closed.In this case current will flow through the three resistances R, X and S.The resistances X and S are in parallel, so their equivalent resistance will be,

$\dfrac{1}{{R'}} = \dfrac{1}{X} + \dfrac{1}{S}$

$\Rightarrow R' = \dfrac{{XS}}{{X + S}}$

The resistances $R'$ and R are in series, so equivalent resistance of the whole circuit will be,

${R_{eq}}^\prime = R' + R$

Putting the value of $R'$we get,

${R_{eq}}^\prime = \dfrac{{XS}}{{X + S}} + R$

Now Current in circuit is I’

Here, current is I^’, voltage is E and resistance is $\dfrac{{XS}}{{X + S}} + R$.

Putting these values in equation (i) we get,

$I^{’} = \dfrac{E}{{\dfrac{{XS}}{{X + S}} + R}}$

$\Rightarrow I^{’} = \dfrac{{E}}{{\dfrac{{XS}}{{X + S}} + R}}$

From the question $\dfrac{I}{2}$ passing through resistance ‘X‘

Now,

$\dfrac{I}{2} = \dfrac{I^{’} \times {S}}{{X+S}}$

$\Rightarrow \dfrac{I}{2} = \dfrac{{E\times {S}}}{XS+RS+RX}$

$\therefore I = \dfrac{{2E \times {S}}}{XS+RS+RX}$

Equating equations (ii) and (iii) we get,

$\dfrac{{2E \times {S}}}{XS+RS+RX} = \dfrac{E}{{R + X}}$

$\Rightarrow \dfrac{{2S}}{{XS + RS + RX}} = \dfrac{1}{{R + X}}$

$\Rightarrow \dfrac{{XS + RS + RX}}{{2S}} = R + X$

$\Rightarrow XS + RS + RX = 2RS + 2SX$

$\Rightarrow RS + SX = RX$

$\Rightarrow RS = \left( {R - S} \right)X$

$\therefore X = \dfrac{{RS}}{{\left( {R - S} \right)}}$

Therefore, the required answer is $X = \dfrac{{RS}}{{\left( {R - S} \right)}}$.

Note: Before calculating the equivalent resistance always check whether the resistances are in series or parallel. For resistances in series, the equivalent resistance is just the addition of the resistances while for resistances in parallel the equivalent resistance is given by the formula, $\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .....$.