Question

When a circuit element $X$ is connected across an a.c. source, a current of$\sqrt 2 A$ flows through it and this current is in phase with the applied voltage. When another element $Y$ is connected across the same a.c. source, the same current flows in the circuit but it leads the voltage by $\dfrac{\pi }{2}radians$. Find the net impedance.

Answer 1

When attempting questions like the ones asked above keep in mind the concepts regarding alternating current and the various terms related to it such as impedance, resistance, current, phase differences and how they affect each other and on what scale.

Complete step-by-step solution:
Before solving this question lets understand a little about Alternating current.
When DC supply voltage is applied to a capacitor, the capacitor is charged slowly and finally it reaches a fully charged position. At this point the charging voltage of a capacitor is equal to the supply voltage. Here the capacitor acts as an energy source as long as voltage is applied. Capacitors don’t allow current $I$ through them after they are fully charged. The current flowing through the circuit depends on the amount of charge in the plates of capacitors and also the current is directly proportional to the rate of change of voltage applied to the circuit.
Now when we talk about AC supply, if it is applied to the capacitor circuit then the capacitor charges and discharges continuously depending on the rate of frequency of supply voltage. The capacitance of a capacitor in AC circuits depends on the frequency of supply voltage applied to it. In AC circuits the capacitors allow current when the supply voltage is continuously changing with respect to time.
From the question we know that the Resistance of circuit element $X$is
$X=R$
From Ohm’s law we also know that
$I=\dfrac{V}{R}$or it can also be written as
$\sqrt{2}=\dfrac{V}{R}$ ……(1)
From the question we also have the values of resistance of $Y$;
Resistance of $Y$, ${{X}_{c}}=\dfrac{1}{wC}$; I=\dfrac{V}{{{X}_{c}}}$$$$\Rightarrow \sqrt{2}=\dfrac{V}{{{X}_{c}}} …….(2)
This implies to us that ${{X}_{c}}=R$ ……(3)
Now when $R$and $C$ are connected in series across the same voltage source, then
Impedance $Z$will be
$\Rightarrow \sqrt{{{R}^{2}}+X_{c}^{2}}$
Which from equation (3) we know;
$\Rightarrow \sqrt{{{R}^{2}}+{{R}^{2}}}$which will be equal to
$\Rightarrow \sqrt{2}R$ ……(4)
$\therefore$Current in the circuit $I$will be
$I=\dfrac{V}{Z}$
$I=\dfrac{V}{\sqrt{2}R}$
Now from equation (1) we know
$\dfrac{V}{R}=\sqrt{2}$
$\therefore I=\dfrac{1}{\sqrt{2}}\times \sqrt{2}=1A$
Hence the net impedance in the circuit will be $1A$.

Note: When we talk about Alternating current often we talk about Capacitive reactance. When we talk about capacitive reactance and frequency we get to know that if frequency is low then reactance is high. The charging current increases with increase in frequency, because the rate of change of voltage increases with time. The reactance is at infinite value where the frequency is zero and it is the same if reversed.