Question

Write down the relationship between root-mean-square value and peak value of an alternating voltage. In the given circuit, find the value of inductive reactance and the potential difference between the ends of the resistance.

Hint:

The root-mean-square value of an alternating voltage is \( \frac{V_{\text{max}}}{\sqrt{2}} \), and the total impedance in an R-L circuit is \( Z = \sqrt{R^2 + X_L^2} \).

Answer 1

Step 1: Relationship Between Root-Mean-Square (rms) Value and Peak Value.
The root-mean-square (rms) value \( V_{\text{rms}} \) of an alternating voltage is related to the peak value \( V_{\text{max}} \) by: \[ V_{\text{rms}} = \frac{V_{\text{max}}}{\sqrt{2}} \] This relation holds for both current and voltage in an AC circuit.
Step 2: Inductive Reactance.
The inductive reactance \( X_L \) is given by the formula: \[ X_L = 2 \pi f L \] where: - \( f \) is the frequency of the AC supply, - \( L \) is the inductance of the coil. Given that \( L = 8.1 \, \text{mH} = 8.1 \times 10^{-3} \, \text{H} \), we can calculate \( X_L \) if the frequency \( f \) is provided.
Step 3: Potential Difference Across the Resistance.
The total impedance \( Z \) in an R-L circuit is given by: \[ Z = \sqrt{R^2 + X_L^2} \] where: - \( R = 10 \, \Omega \) is the resistance, - \( X_L \) is the inductive reactance. The potential difference across the resistance is given by: \[ V_R = I \cdot R \] where \( I \) is the current in the circuit, which can be calculated using the rms voltage and the total impedance: \[ I = \frac{V_{\text{rms}}}{Z} \] Substitute the values to find the potential difference across the resistance.