Question

In circular coil. when no.of turns is doubled & resistance becomes half of the initial then inductance becomes.

• A. $4times$
• B. $2times$
• C. $8times$
• D. No change

Answer 1

The correct option is A

$4times$

Step 1: Given

Initial number of turns= $N$

Final number of turns= $N\text{'}=2N$

Initial resistance=$R$

Final resistance=$R\text{'}=\frac{R}{2}$

Step 2: Examine self-inductance

Self-inductance can be calculated by,

$L=\frac{\mu N^{2}A}{l}$

(Where L is the self-inductance, $\mu$ is the magnetic permittivity, N is the number of turns of the coil, A is the area and l is the length.)

$\Rightarrow L\propto N^2$

So, if the number of turns is doubled,

$$\begin{gathered} L\propto N{\text{'}}^2 \\ \propto {\left(2N\right)}^2 \\ \propto 4N \end{gathered}$$

Therefore, the inductance increases by $4times.$

And as the formula does not contain resistance, self-inductance does not depend on the resistance. Hence, a change in resistance will have no effect.

Therefore, option A is the correct option.