Consider the circuit shown in the figure with input $V(t)$ in volts. The sinusoidal steady state current $I(t)$ flowing through the circuit is shown graphically. The circuit element $Z$ can be ________________.
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If current lags voltage, the element is inductive. If current leads voltage, the element is capacitive.
The input is:
\[
V(t)=\sin(t) = \sin(\omega t), \quad \omega = 1\ \text{rad/s}.
\]
From the graph of $I(t)$, the current reaches its peak at $t=\pi/4$, whereas the voltage $\sin(t)$ reaches its peak at $t=\pi/2$.
Step 1: Determine phase difference.
Voltage peak: $\pi/2$
Current peak: $\pi/4$
Thus the current lags the voltage by:
\[
\phi = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}.
\]
Step 2: Interpret the phase.
Current lagging voltage ⇒ the load is inductive.
Step 3: Compute the reactance.
The current amplitude (from the graph):
\[
I_{\text{max}} = \frac{1}{\sqrt{2}}\ \text{A}.
\]
Voltage amplitude:
\[
V_{\text{max}} = 1\ \text{V}.
\]
Total impedance magnitude:
\[
|Z| = \frac{V_{\text{max}}}{I_{\text{max}}}
= \frac{1}{1/\sqrt{2}} = \sqrt{2}\ \Omega.
\]
Series combination:
\[
Z = 1\Omega + j\omega L.
\]
Magnitude:
\[
\sqrt{1^2 + (\omega L)^2} = \sqrt{2}.
\]
With $\omega = 1$:
\[
1 + L^2 = 2 \implies L^2 = 1 \implies L = 1\ \text{H}.
\]
Step 4: Final conclusion.
The unknown element $Z$ is an inductor of 1 H.
Final Answer: an inductor of 1 H
