Question

An ac source producing emf $ \text{e}={{\text{E}}_{0}}\left[ \cos \left( 100\text{ }\!\!\pi\!\!\text{ t} \right)+\cos \left( 500\text{ }\!\!\pi\!\!\text{ t} \right) \right] $ is connected in series with a capacitor and a resistor. The steady state current in the circuit is found to be $ \text{i}={{\text{I}}_{1}}\cos \left( 100\text{ }\!\!\pi\!\!\text{ t}+{{\text{ }\!\!\theta\!\!\text{ }}_{1}} \right)+{{\text{I}}_{2}}\cos \left( 500\text{ }\!\!\pi\!\!\text{ t}+\text{ }\!\!\theta\!\!\text{ 2} \right) $
(A) $ {{\text{I}}_{1}}>{{\text{I}}_{2}} $
(B) $ {{\text{I}}_{1}}\text{=}{{\text{I}}_{2}} $
(C) $ {{\text{I}}_{1}}\text{}{{\text{I}}_{2}} $
(D) Nothing can be said

Answer 1

Hint: A capacitor is a passive device that stores energy in its electric field and returns energy to the circuit whenever required capacitors do not store Ac voltage
Charge on capacitor is Q=CV
and current $ \text{i}=\dfrac{\text{dQ}}{\text{dt}} $

Complete step by step solution
We can assume two different sources connected in series with same orientation
$ \begin{align}
 & {{\text{E}}_{1}}={{\text{E}}_{\text{0}}}\cos \text{ 100 }\!\!\pi\!\!\text{ t} \\
 & {{\text{E}}_{2}}={{\text{E}}_{\text{0}}}\cos \text{ 500 }\!\!\pi\!\!\text{ t} \\
\end{align} $
Such that
$ \begin{align}
 & \text{e}={{\text{E}}_{\text{0}}}\cos \text{ 100 }\!\!\pi\!\!\text{ t}+{{\text{E}}_{\text{0}}}\cos \text{ 500 }\!\!\pi\!\!\text{ t} \\
 & \text{ =}{{\text{E}}_{0}}\left[ \cos \text{ 100 }\!\!\pi\!\!\text{ t}+\cos \text{ 500 }\!\!\pi\!\!\text{ t} \right] \\
\end{align} $
Becomes,
$ \text{e}={{\text{E}}_{1}}+{{\text{E}}_{2}} $
The charge on the capacitor during the steady state is given by
Q=CE
Putting the value of E in this equation
 $ \text{Q}=\text{C }{{\text{E}}_{0}}\left[ \cos \text{ 100 }\!\!\pi\!\!\text{ t}+\cos \text{ 500 }\!\!\pi\!\!\text{ t} \right] $
The steady state current is, thus given by
$ \begin{align}
 & \text{i}=\dfrac{\text{dQ}}{\text{dt}} \\
 & \text{i}=\dfrac{\text{d}}{\text{dt}}\left[ \text{C }{{\text{E}}_{0}}\left[ \cos \text{ 100 }\!\!\pi\!\!\text{ t}+\cos \text{ 500 }\!\!\pi\!\!\text{ t} \right] \right] \\
 & \text{i}=\dfrac{\text{d}}{\text{dt}}\left( \text{C }{{\text{E}}_{0}}\cos \text{ 100 }\!\!\pi\!\!\text{ t} \right)+\dfrac{\text{d}}{\text{dt}}\left( \text{C }{{\text{E}}_{0}}\cos \text{ 500 }\!\!\pi\!\!\text{ t} \right) \\
 & \text{C }{{\text{E}}_{0}}\left( 100\text{ }\!\!\pi\!\!\text{ } \right)\text{ sin100 }\!\!\pi\!\!\text{ t}+\text{C }{{\text{E}}_{0}}\left( 500\text{ }\!\!\pi\!\!\text{ } \right)\text{ sin500 }\!\!\pi\!\!\text{ t} \\
 & \text{ }\left[ \because \dfrac{\text{d}}{\text{dt}}\cos \text{ t}=\sin \text{ t} \right] \\
\end{align} $
$ {{\text{i}}_{1}}\text{ sin 100 }\!\!\pi\!\!\text{ t}+{{\text{i}}_{2}}\text{ sin 500 }\!\!\pi\!\!\text{ t} $
Where $ {{\text{i}}_{1}}=\text{C }{{\text{E}}_{0}}\left( \text{100 }\!\!\pi\!\!\text{ t} \right)\text{ } $
$ {{\text{i}}_{2}}=\text{C }{{\text{E}}_{0}}\left( \text{500 }\!\!\pi\!\!\text{ t} \right)\text{ } $
Now, using the principle of superposition
We get,
 $ \text{i}={{\text{i}}_{1}}\cos \text{ }\left( \text{100 }\!\!\pi\!\!\text{ t}+{{\text{ }\!\!\theta\!\!\text{ }}_{1}} \right)+{{\text{i}}_{2}}\cos \text{ }\left( \text{500 }\!\!\pi\!\!\text{ t}+{{\text{ }\!\!\theta\!\!\text{ }}_{2}} \right)\text{ } $
From this expression, we get
$ \begin{align}
 & {{\text{i}}_{1}}={{\text{E}}_{0}}\text{100 }\!\!\pi\!\!\text{ C} \\
 & {{\text{i}}_{2}}={{\text{E}}_{0}}\text{500 }\!\!\pi\!\!\text{ C} \\
\end{align} $
From this we can easily predict that
$ {{\text{i}}_{2}}>{{\text{i}}_{1}} $
Therefore option (C) is correct.

Note
Capacitor and battery are different devices because the potential energy in a capacitor is stored in an electric field, where a battery stores its potential energy in a chemical form. Knowledge of some trigonometric rules and differentiation rule should be must be careful while solving the differentiation because students get confused easily