Question

A pure resistive circuit element $X$ when connected to an AC supply of peak voltage $200\,{\text{V}}$ gives a peak current of $5\,{\text{A}}$ . A second current element $Y$ when connected to the same AC supply gives the same value of peak current but the current lags behind by $90^\circ$ . If the series combination of $X$ and $Y$ is connected to the same supply, what is the impedance of the circuit?
A. $40\,\Omega$
B. $80\,\Omega$
C. $40\sqrt 2 \,\Omega$
D. $2\sqrt {40} \,\Omega$

Answer 1

First of all, we will calculate the individual resistances of the elements. We know that for a resistor, the current and the voltage are in phase. For an inductor, the voltage leads the current. After that we will find the net impedance.

Complete step by step solution:
In the given question, we are supplied with the following data:
When we connect a circuit element $X$ to an AC supply of peak voltage $200\,{\text{V}}$ we get a peak current of $5\,{\text{A}}$ .A second current element $Y$ when connected to the same AC supply, we get the same peak current, but in this case, the current lags behind by $90^\circ$ .We are asked to find the impedance of the circuit, if the series combination of $X$ and $Y$ is connected to the same supply.

To begin with, we know a current is in phase with a voltage, so the element $X$ that we have connected in the circuit, must be definitely a resistor.
So, we use ohm’s law to find the resistance of the resistor:
$R = \dfrac{{{V_0}}}{{{I_0}}}$ …… (1)
Where,
$R$ indicates the resistance of the resistor.
${V_0}$ indicates the voltage of the AC supply.
${I_0}$ indicates the peak current.
Now we substitute the required values in the equation (1) and we get:
R = \dfrac{{{V_0}}}{{{I_0}}} \\\ \Rightarrow R = \dfrac{{200}}{5} \\\ \Rightarrow R = 40\,\Omega \\\
Therefore, the resistance of the resistor is found to be $40\,\Omega$ .
Again, we know that, in an inductive circuit, the current lags behind the voltage, so after connection of element $Y$ , this happens.
So, the element $Y$ must be an inductor.
We can calculate the impedance of the inductor by using the formula:
${X_{\text{L}}} = \dfrac{{{V_0}}}{{{I_0}}}$ …… (2)
Where,
${X_{\text{L}}}$ indicates the impedance of the inductor.
${V_0}$ indicates the voltage of the AC supply.
${I_0}$ indicates the peak current.
Now we substitute the required values in the equation (2) and we get:
{X_{\text{L}}} = \dfrac{{{V_0}}}{{{I_0}}} \\\ \Rightarrow {X_{\text{L}}} = \dfrac{{200}}{5} \\\ \Rightarrow {X_{\text{L}}} = 40\,\Omega \\\
Therefore, the impedance of the inductor is found to be $40\,\Omega$ .
As the two elements are connected in series. The net impedance provided by the circuit is:
$z = \sqrt {{R^2} + X_{\text{L}}^2} \\\ \Rightarrow z = \sqrt {{{40}^2} + {{40}^2}} \\\ \therefore z = 40\sqrt 2 \,\Omega$
Hence, the net impedance provided by the circuit is $40\sqrt 2 \,\Omega$ .

The correct option is C.

Note: While solving this problem, many students seem to have confusion regarding the lag, lead and phase. It is important to remember that in an inductive circuit, voltage leads the current while in a capacitive circuit, current leads the voltage. It is very essential to know this, so that we can correctly determine the type of the element.