Question

An AC source generating a voltage \[V = {V_m}\sin (\omega t)\] is connected to a capacitor of capacitance C. Find the expression of the current i flowing through it. Plot a graph of V and i versus \[\omega t\] to show that the current is π/2 ahead of the voltage.

Answer 1

Hint: Understand the concept of current as flow rate of charge per unit time. Also use the formula for voltage passing through a capacitor of capacitance C to deduce the expression. Draw the plot for different values of \[\omega \].

Complete answer:
We know that there is an AC source generating a voltage of \[V = {V_m}\sin (\omega t)\], that is attached to a capacitor of capacitance C.
Now, we know that the voltage passing through the capacitor V is equal to \[V = Q/C\], where Q is the charge value in coulombs and C is capacitance value in farads.
Now, current I in any circuit is defined as the flow rate of charges per unit time. Mathematically,
\[i = \dfrac{{dQ}}{{dt}}\]
Now, rearranging the Voltage inside a capacitor equation we get,
\[V \times C = Q\]
Substituting this in the current equation we get,
\[i = \dfrac{{d[V \times C]}}{{dt}}\]
Now, V is given as \[{V_m}\sin (\omega t)\]for the given circuit, substituting that in the above equation we get\[i = \dfrac{{d[C \times {V_m}\sin (\omega t)]}}{{dt}}\]
Applying differentiation inside, we get,
\[i = C \times \dfrac{{d[{V_m}]}}{{dt}} \times \omega \cos (\omega t)\]
Note: Differentiation of \[\sin (\omega t)\]is \[\cos (\omega t)\]multiplied by differentiation of \[\omega t\], which is \[\omega \].
Now, \[\cos (\omega t)\]can also be written as \[\sin (\omega t + \pi /2)\], due to quadrants.
\[i = C \times \dfrac{{d[{V_m}]}}{{dt}} \times \omega \times \sin (\omega t + \pi /2)\]
Taking constant term C and\[\omega \]to the denominator,
\[i = \dfrac{{\dfrac{{d[{V_m}]}}{{dt}} \times \sin (\omega t + \pi /2)}}{{1/C \times \omega }}\]
Now, we know that the instantaneous current passing through a capacitor of capacitance C is given as ,
\[{i_m} = C \times \dfrac{{d[{V_m}]}}{{dt}}\]
Applying this concept to the above equation,
\[ \Rightarrow i = {i_m}\sin (\omega t + \pi /2)\]
Where \[{i_m} = \dfrac{{{V_m}}}{{1/C\omega }}\]
The term \[1/C\omega \] is called capacitive reactance and is represented as \[{X_c}\].
Now, plot the graph for different values of \[\omega t\] ranging from zero to\[{360^ \circ }\].
The resulting graph will see that the current is leading by a phase difference of \[\pi /2\].

Thus the expression is derived and the graph is drawn accordingly.

Note:
Capacitive reactance is defined as the total opposition to the current flow offered by a capacitor of capacitance C. It is similar to resistance but it’s not equal to resistance. It is represented by the term\[{X_c}\].