Question

In the circuit shown the ammeter A reads a current of I₁ amp. 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 I₂ amp. Then –

• A. I₁> I₂
• B. I₁ = I₂
• C. I₁< I₂
• D. The relation between I₁ and I₂ will depend upon the relative value of resistances R₁, R₂ and R₃

Answer 1

Answer: (B)

I₁ = I₂

Answer 2

I₁ = $\frac{R_{2}I}{R_{2} + R_{3}}$

= $\frac{R_{2}}{R_{2} + R_{3}}$ × $\frac{E}{\left\lbrack R_{1} + \frac{R_{2}R_{3}}{R_{2} + R_{3}} \right\rbrack}$

= $\frac{ER_{2}}{R_{1}R_{2} + R_{2}R_{3} + R_{3}R_{1}}$

After interchanging ammeter and source of e.m.f., new current

⇒ I₂ = $\frac{R_{2}I}{R_{1} + R_{2}}$

= $\frac{E}{\left\lbrack R_{3} + \frac{R_{1}R_{2}}{R_{1} + R_{2}} \right\rbrack}$

= $\frac{ER_{2}}{R_{1}R_{2} + R_{2}R_{3} + R_{3}R_{1}}$

= I₁