Question

A 220 V, 50 Hz a.c generator is connected to an inductor and $50\;\Omega$ resistance in series. The current in the circuit is 1.0 A. The P.D across the inductor is:
A. 102.2 V
B. 186.4 V
C. 214. 24 V
D. 302 V

Answer 1

The given problem can be resolved using the concepts as well as the fundamental relation between the current and impedance in an inductive circuit. Moreover, the mathematical relation for the impedance for inductors, in terms of resistance can also be used.

Complete step by step answer:
Given:
The voltage supply is, V = 220 V.
The frequency is, f = 50 Hz.
The resistance is, $R = 50\;\Omega$.
The current in the circuit is, $I = 1.0\;{\rm{A}}$.
The expression for the current is,
$I = \dfrac{V}{{\sqrt {X_L^2 + {R^2}} }}$
Here, ${X_L}$ is the inductive resistance.
Solving as,

1\;{\rm{A}} = \dfrac{{200\;{\rm{V}}}}{{\sqrt {X_L^2 + {{\left( {50\;\Omega } \right)}^2}} }}\\\ \sqrt {X_L^2 + {{\left( {50\;\Omega } \right)}^2}} = 200\;{\rm{V}}\\\ {{\rm{X}}_L} = 214.24\;\Omega \end{array}$$ The potential difference is, $$\begin{array}{l} {V_L} = I \times {X_L}\\\ {V_L} = 1\;{\rm{A}} \times 214.24\;\Omega \\\ {V_L} = 214.24\;{\rm{V}} \end{array}$$ Therefore, the required potential difference is of 214.24 V **So, the correct answer is “Option C”.** **Note:** To resolve the given problem, the concept of resistive as well as the inductive circuits is to be remembered, along with basic mathematical relation of Inductive resistance with voltage supply and Impedance. Moreover, the correct relation for each of the circuits is to be determined, to obtain the appropriate results for the given set of conditions.