Question

A capacitor of capacitance \[1\mu F\] is charged to a potential of 1 V. It is connected in parallel to an indicator of inductance \[{{10}^{-3}}H\]. The maximum current that will flow in the circuit has the value.
A. \[\sqrt{1000}mA\]
B. \[1mA\]
C. \[1\mu A\]
D. \[1000mA\]

Answer 1

Hint: In this question, we have been asked to calculate the maximum current in the given LC circuit. From the formula for LC oscillator, we know that maximum charge is given as maximum current over oscillation frequency. Therefore, we will calculate the value of charge as it is given as the product of capacitance and voltage. We will also calculate the oscillation frequency and thus calculate the maximum current.
Formula Used:
\[Q=CV\]
\[\omega =\dfrac{1}{\sqrt{LC}}\]
\[Q=\dfrac{{{I}_{m}}}{\omega }\]

Complete answer:
We know that charge in a conductor can be given by
\[Q=CV\]
 

After substituting the given values
We get,
\[Q=1\times {{10}^{-6}}\times 1\]
Therefore,
\[Q=1\times {{10}^{-6}}C\]
We know that given circuit is LC circuit. Therefore, from the LC oscillator formula we know,
\[Q=\dfrac{{{I}_{m}}}{\omega }\] ……………. (1)
Now, we know that \[\omega \] for LC circuit is given by
\[\omega =\dfrac{1}{\sqrt{LC}}\]
After substituting the given values
We get,
\[\omega =\dfrac{1}{\sqrt{{{10}^{-3}}\times {{10}^{-6}}}}\]
Therefore,
\[\omega =\dfrac{1}{\sqrt{{{10}^{-9}}}}\]
After substituting the values in equation (1)
 We get,
\[{{10}^{-6}}=\dfrac{{{I}_{m}}}{\dfrac{1}{\sqrt{{{10}^{-9}}}}}\]
Therefore,
\[{{I}_{m}}=\sqrt{1000}\times {{10}^{-3}}A\]
It can be written as
\[{{I}_{m}}=\sqrt{1000}mA\]

Therefore, the correct answer is option A.

Note:
An inductor is a two terminal electrical device that is used to store energy in a magnetic field when the current flows through it. A capacitor is a two terminal electrical device which also stores energy but it stores it in an electric field. The inductor resists the electric current flowing through it, while the capacitor resists the change in voltage applied across its terminals.