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
$\Rightarrow I_2$ $=\frac{R_2}{R_1+R_2}\frac{E}{\left[R_3+\frac{R_1{R}_2}{R_1+R_2}\right]}$
$\frac{E{R}_2}{R_1{R}_2+R_2{R}_3+R_3{R}_1}=I_1$