A wire of 20 Ω resistance is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 4.0 volt battery. Find the current drawn from the battery.

Solution

When, any resistor is stretched to double its original length, the new resistance becomes four times of its original resistance because

R \propto \frac{1}{A} i . e ., R \propto \frac{1}{π {(\frac{d}{2})}^{2}}

Here,

Resistance of wire, R = 20 Ω Potential applied across the circuit, V = 4 V

∴

New resistance will be 4R.

4 \times 20 = 80 Ω [R \propto \frac{4}{{πd}^{2}}]

As the length is cut into two equal parts, resistance of each part is

\frac{80}{2} = 40 Ω

R_{1} = 40 Ω, R_{2} = 40 Ω

Effective resitance in parallel combination

R_{P}

is given by,

\frac{1}{R_{P}} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20}

[∵ \frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}]

∴ R_{P} = 20 Ω

Current drawn from the battery, I =

\frac{V}{R_{P}} = \frac{4.0}{20} = 0.2 A .

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