Question

In the circuit shown the ideal ammeter $A$ reads a current of $I_{1} \, A$ . Now the source of e.m.f. and the ammeter are physically interchanged, i.e. the source is put between $B$ and $C$ and the ammeter between $A$ and $B$ . The ammeter now reads a current of $I_{2} \, A$ . Then -

• A. $I_{1}>I_{2}$
• B. $I_{1}=I_{2}$
• C. $I_{1} < I_{2}$
• D. The relation between $I_{1}$ and $I_{2}$ will depend upon the relative value of resistances $R_{1}, \, R_{2}$ and $R_{3}$

Answer 1

Answer: (B)

$I_{1} \, $ $=\frac{R_{2}}{R_{2} + R_{3}}\times \frac{E}{\left[R_{1} + \frac{R_{2} R_{3}}{R_{2} + R_{3}}\right]}$
$= \, \frac{E R_{2}}{R_{1} R_{2} + R_{2} R_{3} + R_{3} R_{1}}$
After interchanging ammeter and source of e.m.f., new current
$\Rightarrow I_{2}$ $= \, \frac{R_{2}}{R_{1} + R_{2}}\frac{E}{\left[R_{3} + \frac{R_{1} R_{2}}{R_{1} + R_{2}}\right]}$
$\frac{E R_{2}}{R_{1} R_{2} + R_{2} R_{3} + R_{3} R_{1}}=I_{1}$