91Ó°ÊÓ

1. Emf of the ac generator ${∈}_m=30.0\mathrm{V}$
2. Series capacitance $C=1.5\mathrm{μ¹ó}=1.5×{10}^{-6}\mathrm{F}$

As it passes through a capacitor, the obstruction offered in the path of alternating current is known as capacitive reactance. By determining the reactance due to the oscillation frequency within a capacitor, we can get the current value by substituting this value in equation of Ohm's law.

The capacitive reactance of a capacitor,

$X_c=\frac{1}{2\mathrm{Ï€´Ú°ä}}$…..(¾±)

The current equation from Ohm’s law,

$i_c=\frac{{∈}_m}{X_c}$…..(¾±¾±)

Here, $f$is the frequency of oscillations, $C$is the capacitance of the capacitor, and${∈}_m$ is the emf applied across the circuit.

For frequency $f=1\mathrm{kHz}$, The capacitive reactance is given using equation (i) as follows:

$$\begin{gathered} X_c=\frac{1}{2\mathrm{Ï€}×\left(1×{10}^3\right)\mathrm{Hz}×\left(1.5×{10}^{-6}\right)\mathrm{F}} \\ {X}_c=106.1\mathrm{Î\copyright } \end{gathered}$$

Thus, using this above value in equation (ii), the value of the amplitude of the resulting alternating current is given as:

$$\begin{gathered} i_c=\frac{30.0\mathrm{V}}{0.283\mathrm{Î\copyright }} \\ {i}_c=0.283\mathrm{A} \end{gathered}$$

Hence, the value of the current is$0.283\mathrm{A}$.

For frequency $f=8\mathrm{kHz}$, the capacitive reactance is given using equation (i) as follows:

$$\begin{gathered} X_c=\frac{1}{2\mathrm{Ï€}×\left(8×{10}^3\right)\mathrm{Hz}×\left(1.5×{10}^{-6}\right)\mathrm{F}} \\ {X}_c=13.26\mathrm{Î\copyright } \end{gathered}$$

Thus, using this above value in equation (ii), the value of the amplitude of the resulting alternating current is given as:

$$\begin{gathered} i_c=\frac{30.0\mathrm{V}}{13.26\mathrm{Î\copyright }} \\ {i}_c=2.26\mathrm{A} \end{gathered}$$

Hence, the value of the current is$2.26\mathrm{A}$.