In the part of an a.c. circuit as shown, the resistance $R = 0.2\ \Omega$. At a certain instant $(V_A - V_B) = 0.5\text{ V}$, $I = 0.5\text{ A}$ and $\frac{\Delta I}{\Delta t} = 8\text{ A/s}$. The inductance of the coil is
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When current is increasing ($\frac{dI}{dt} > 0$), an inductor acts like a small battery opposing the flow, which creates an additional voltage drop along your path. Subtracting the quick resistive drop ($0.5 \times 0.2 = 0.1\text{ V}$) from the total voltage ($0.5\text{ V}$) leaves exactly $0.4\text{ V}$ across the inductor. Dividing $0.4$ by $8$ gives $0.05\text{ H}$ instantly.
Step 1: Picture the branch. Between nodes $A$ and $B$ a resistor $R = 0.2\,\Omega$ and a coil of inductance $L$ sit in series. The instant values are $V_A - V_B = 0.5\,\text{V}$, $I = 0.5\,\text{A}$ and $\dfrac{\Delta I}{\Delta t} = 8\,\text{A/s}$. Step 2: Write the two voltage drops. Across the resistor the drop is $IR$, and across the inductor it is $L\dfrac{\Delta I}{\Delta t}$. Step 3: Apply Kirchhoff along the branch. Moving from $A$ to $B$ with the current, \[ V_A - V_B = IR + L\frac{\Delta I}{\Delta t}. \] Step 4: Put in the numbers. \[ 0.5 = (0.5)(0.2) + L(8) = 0.1 + 8L. \] Step 5: Isolate the inductive term. \[ 8L = 0.5 - 0.1 = 0.4. \] Step 6: Solve for $L$. \[ L = \frac{0.4}{8} = 0.05\,\text{H}. \] That is option (D). Notice the resistor only ate $0.1\,\text{V}$, leaving $0.4\,\text{V}$ for the coil. \[ \boxed{L = 0.05\,\text{H}} \]
