Question

In an AC circuit containing only an inductor, the current [H] [width=0.5\linewidth]{113.png}

• A. [H] [width=0.25\linewidth]{113a.png}
• B. [H] [width=0.5\linewidth]{113b.png}
• C. [H] [width=0.25\linewidth]{c.png}
• D. [H] [width=0.25\linewidth]{d.png}

Hint:

Remember that for a purely inductive circuit, the phase difference between voltage and current is always $\frac{\pi}{2}$ radians or 90 degrees.

Answer 1

The Correct Option is A

Step 1: Concept
In an AC circuit, the current through an inductor lags behind the applied voltage by a phase angle of $\frac{\pi}{2}$ radians.

Step 2: Meaning
This means that if the voltage across an inductor is at its peak value, the current will be zero. Conversely, when the current reaches its maximum value, the voltage will be zero.

Step 3: Analysis
To understand why this happens, consider the relationship between the voltage and current in an inductor. The voltage $V$ across an inductor is given by: \[V = L \frac{dI}{dt}\] where $L$ is the inductance of the inductor and $\frac{dI}{dt}$ is the rate of change of current with respect to time. If we assume a sinusoidal voltage source, the voltage can be represented as: \[V(t) = V_0 \sin(\omega t)\] where $V_0$ is the peak voltage and $\omega$ is the angular frequency. The current through the inductor will then be: \[I(t) = \frac{1}{\omega L} \cos(\omega t) + C\] For simplicity, we can assume that at time $t=0$, the current $I(0)=0$. This gives us the solution for the current as: \[I(t) = \frac{V_0}{\omega L} \sin\left(\omega t - \frac{\pi}{2}\right)\] This shows that the current lags behind the voltage by $\frac{\pi}{2}$ radians.

Step 4: Conclusion
The phase difference between the voltage and current in an inductor is $\frac{\pi}{2}$ radians, which means the current lags behind the voltage by this angle. Final Answer: (A)