Name: Superconductivity: An Overview
Uploaded: 2019-07-19
Duration: 2 min 51 s
Description: Resistance of mercury at very low temperatures reduces to nearly zero. Certain materials exhibit this property of superconductivity. Let's learn more.

Question

A potentiometer wire carries a steady current. The potential difference across

70 cm

length of it balances the potential difference across a

2 Ω

coil supplied by a cell of emf

2.0 V

and an unknown internal resistance

r

. When a

1 Ω

coil is placed in parallel with the

2 Ω

coil, a length equal to

50 cm

of the potentiometer wire is required to balance the potential difference across the parallel combination. Find the value of

r

.

Accepted Answer

When a $1 Ω$ coil is placed in parallel with the $2 Ω$ coil, a length equal to $50 cm$ of the potentiometer wire is required to balance the potential difference across the parallel combination.

In first case. Current sent by the $2.0 V$ cell through $2 Ω$ coil,

$I_{1} = \frac{ξ}{Total resistance} = \frac{2}{2 + r}$

Potential drop across $2 Ω$ coil,

$V_{1} = R I_{1} = 2 \times \frac{2}{2 + r} = \frac{4}{2 + r}$

But $V_{1} \propto 70 cm$

$∴ \frac{4}{2 + r} \propto 70$ .............(i)

In second case. The combined resistance of the parallel combination of $2 Ω$ and $1 Ω$ coil,

$R_{p} = \frac{2 \times 1}{2 + 1} = \frac{2}{3} Ω$

Current sent by the cell through the parallel combination,

$I_{2} = \frac{E}{Total resistance} = \frac{2}{(2 / 3) + r} = \frac{6}{2 + 3 r}$

Potential drop across $R_{p}$ ,

$V_{2} = R_{p} I_{2} = \frac{2}{3} \times \frac{6}{2 + 3 r} = \frac{4}{2 + 3 r}$

But $V_{2} \propto 50 cm$

$\frac{4}{2 + 3 r} \propto 50$ .................(ii)

Dividing (i) by (ii), we get

$\frac{4}{2 + r} \times \frac{2 + 3 r}{4} = \frac{70}{50}$

or $r = 0.5 Ω$ .

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