Question

A Capacitor of $2\mu F$ is connected in a radio circuit. The source frequency is $1000\,Hz$. If the current through the capacitor branch is $2mA$ then voltage across the capacitor is
A. 0.16 V
B. 0.32 V
C. 156 V
D. 79.5 V

Answer 1

In the question above capacitance , frequency and current has been given. By ohm’s law we know that voltage is directly proportional to current and $V = iR$. In this case. First we find the capacitive reactance ${\chi _c}$ and then we find the voltage $V$.

Formula used:
$V = i\dfrac{1}{{\omega C}}$
where ${\chi _c}$=capacitive reactance.

Complete step by step answer:
We know that AC is alternating current which periodically reverses its direction unlike DC. AC is generally used for distribution of power .By ohm’s law we know that,
$V=IR$…………....(I=current, V=voltage ,R=resistor)

As this is an AC circuit the voltage will be given by $V = i{\chi _c}$.Capacitive reactance Is the opposition offered by capacitor to the change in voltage across the element. Look at the diagram below :

We are suppose to find the voltage ;
We know that:
${\chi _c} = \dfrac{1}{{\omega C}}$ (where $\omega =$frequency)
Substituting the values of C, $\omega$ and we get:
$V = \dfrac{{2 \times {{10}^{ - 2}}}}{{2\pi \times 1000 \times 2 \times {{10}^{ - 6}}}}$
$V = 0.16\,V$
Hence after solving we see that voltage =0.16 V.

Therefore the correct answer is option A.

Additional information: For a purely capacitive AC circuit, instantaneous value of current is given by: $i = \left( {\dfrac{{{V_0}}}{{\dfrac{1}{{\omega C}}}}} \right)\cos \omega t$
$i = {i_0}\sin \left( {\omega t + \phi } \right)$. This equation tells us that the current leads voltage by $\left( {\dfrac{\pi }{2}} \right)$.

Note: The capacitive reactance is given by ${\chi _c} = \dfrac{1}{{\omega C}}$ and inductive reactance is given by ${\chi _L} = \omega L$.do not get confused with the formulas. In a pure capacitive circuit current always leads the voltage by ${90^ \circ }$ and average power applied to it is zero. The capacitive reactance increases with decrease in frequency.