Question

In circular coil. when no. of turns is doubled \& resistance becomes half of the initial then inductance becomes ...... (a) 4 times (b) 2 times (c) 8 times (d) No change

Short answer

The inductance becomes \(2\) times the initial inductance when the number of turns is doubled and resistance becomes half of the initial.

Step-by-step solution

Determine the initial inductance of the coil

Let's denote the initial number of turns as N, resistance as R, and inductance as L. At first, we haven't made any changes to these values.

Find the formula to connect inductance, number of turns, and resistance

From the formula of the inductor, we know that inductance(L) is proportional to the square of the number of turns (N^2) and the reluctance R (L∝N^2R). So, we can write: L = k × N^2 × R where k is a proportionality constant.

Double the number of turns and halve the resistance

Now, let's double the number of turns and halve the resistance. We denote the new number of turns as N' and the new resistance as R'. Thus, N' = 2N and R' = R/2.

Find the new inductance L'

Using the formula from Step 2, we can find the new inductance L' with the new values of N' and R'. L' = k × (N')^2 × R' Substituting values of N' and R' we get: L' = k × (2N)^2 × (R/2)

Simplifying the equation and finding the ratio of the new inductance to the initial inductance

We can simplify the equation for the new inductance and find the ratio of L' to L. L' = k × 4N^2 × R/2 L' = 2k × N^2 × R Now, let's find L'/L: L'/L = (2k × N^2 × R) / (k × N^2 × R) The factors of k, N^2, and R cancel out: L'/L = 2

Choose the correct answer

From our calculations in Step 5, we've determined that the new inductance L' is 2 times the initial inductance L. So, the correct answer is: (b) 2 times