Question

An alternating voltage \[E=200\sqrt{2}\sin (100t)\] volt is connected to $1\text{ }\!\!\mu\!\!\text{ f}$ capacitor through a.c. ammeter. The reading of ammeter shall be
A. 10mA
B. 20mA
C. 40 mA
D. 80 mA

Answer 1

Hint: Firstly we should have a basic knowledge of A.C circuit containing capacitor and effect of capacitor in an a.c circuit.We will find \[{{E}_{rms}}\text{ and }{{X}_{C}}\] and then by using formula for \[{{I}_{rms}}\] we find the answer.

Formula used:
\[\begin{align}
  & E={{E}_{0}}sin(\omega t) \\
 & {{X}_{C}}=\dfrac{1}{\omega C} \\
\end{align}\]

Complete step by step answer:
Firstly we would find the impedance

We know that

\[E={{E}_{0}}sin(\omega t)\]
Now we will compare given equation with the equation \[E=200\sqrt{2}\sin (100t)\]
Hence,
\[~{{E}_{rms}}=\dfrac{{{E}_{0}}}{\sqrt{2}}=200V\text{ and }~\omega =100{{s}^{-1}}\]

Now impedance of the given circuit is

\[\begin{align}
  & E=200\sqrt{2}sin(100t) \\
 & E={{E}_{0}}sin(\omega t) \\
& {{E}_{rms}}=\dfrac{{{E}_{0}}}{\sqrt{2}}=200V~\text{ }~\text{ }~\text{ }~\text{ }~and~\text{ }~\text{ }~\text{ }~\text{ }~\text{ }~\omega =100{{s}^{-1}} \\
 & ~{{X}_{C}}=\dfrac{1}{\omega C}\Rightarrow {{X}_{C}}=\dfrac{1}{100\times {{10}^{-6}}}={{10}^{4}} \\
\end{align}\]

Now we know the formula of \[{{I}_{rms}}\]

\[\begin{align}
  & Hence,~{{I}_{rms}}=\dfrac{{{E}_{rms}}}{{{X}_{C}}}=\dfrac{200}{{{10}^{4}}}A \\
 & {{I}_{rms}}=20mA \\
\end{align}\]

Hence by solving we get the reading of the ammeter should be 20mA.

Hence, the correct option is (B).

Additional Information:
Let us see why A.C source is used for power distribution. Most large power-distribution systems are AC. Moreover, the facility is transmitted at much higher voltages than the 120-V AC (240 V in most parts of the world) we use in homes and on the work. Economies of scale make it cheaper to create a couple of very large electric power-generation plants than to create numerous small ones. This necessitates sending power long distances, and it's obviously important that energy losses on the way be minimized. High voltages are often transmitted with much smaller power losses than low voltages. The crucial factor is that it's much easier to extend and reduce AC voltages than DC, so AC is employed in most large power distribution systems.

Note:
The capacitive reactance varies inversely with the frequency. As if increases, \[{{X}_{C}}\] decreases. We know that as the frequency across the capacitor decreases its reactance value increases. This variation is called the capacitor’s complex impedance.