Question

An alternating voltage is concerned in series with a resistance $r$ and inductance $L$. If the potential drop across the resistance is $200$ volt and across the inductance is $150$ volt. The applied voltage is:
A. $350\text{ V}$
B. $250\text{ V}$
C. $500\text{ V}$
D. $300\text{ V}$

Answer 1

To solve these types of questions we need to know about the $RL$ circuit. A $RL$ circuit is an electric circuit that is made up of resistors and inductors. In the question it is mentioned that an alternating voltage is concerned in series with a resistance $r$ and inductance $L$, hence the same current will flow through the inductor and resistor.

Formula used:
$V=\sqrt{{{\left( {{V}_{R}} \right)}^{2}}+{{\left( {{V}_{L}} \right)}^{2}}}$
Here $V$ is the applied voltage,
${{V}_{R}}$ is the potential drop across the resistance of the circuit,
${{V}_{L}}$ is the potential drop across the inductance.

Complete step-by-step solution:
Let us assume that current $''i''$ flows through the circuit. Since the resistance $r$ and inductance $L$ are connected in series hence the same current will flow through them. The potential difference across the inductor $\left( {{V}_{L}} \right)$ leads the current by $\dfrac{\pi }{2}$ while the potential difference across the resistor $\left( {{V}_{R}} \right)$ will be in phase with the current, hence the resultant potential difference i.e., the applied potential difference will be as follows:
$V=\sqrt{{{\left( {{V}_{R}} \right)}^{2}}+{{\left( {{V}_{L}} \right)}^{2}}}$
It is given in the question that the potential drop across the resistance is $200$ volt and across the inductance is $150$ volt, thus on substituting the values in the equation, the applied voltage will be:
$\begin{aligned} & V=\sqrt{{{\left( 200 \right)}^{2}}+{{\left( 150 \right)}^{2}}} \\\ & \Rightarrow V=\sqrt{40000+22500} \\\ & \Rightarrow V=\sqrt{62500} \\\ & \therefore V=250\text{ V} \\\ \end{aligned}$
Hence, the applied voltage will be $250\text{ V}$ and the correct option will be $B$.

Note: To solve these types of questions, we need to remember the phase difference between the potential difference and current in various electrical components of the circuit. It must be remembered that in a $RL$ circuit the potential difference across the inductor $\left( {{V}_{L}} \right)$ leads the current by $\dfrac{\pi }{2}$, across the resistor $\left( {{V}_{R}} \right)$ is in phase with the current.