Question 66

A 8Ω resistance wire is doubled on it. Calculate the new resistance of the wire.

When the wire is doubled the length will be halved and area will be double or twice

$$R = \frac{P\times L}{A}$$

$$R1 = \frac{P\times \frac{L}{2}}{2A}$$ , where R1 is the new resistance

$$\frac{R1}{R}=\frac{\frac{P\times \frac{L}{2}}{2A}}{\frac{P\times L}{A}}$$

$$\frac{R1}{R} = \frac{1}{4}$$

$$R1 = \frac{R}{4}$$

Putting the value of R=8 Ω,

$$R1=\frac{8}{4} = 2Ω$$

Option D is correct.

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