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: Understanding the Question:
The problem presents a branch of an alternating current circuit containing a resistor $R$ and an inductor coil $L$ connected in series between two network nodes, $A$ and $B$.
At a specific instant, the potential difference across the branch is $(V_A - V_B) = 0.5\text{ V}$, the current flowing through it is $I = 0.5\text{ A}$, and the current is increasing at a rate of $\frac{dI}{dt} = 8\text{ A/s}$. We need to calculate the value of the self-inductance $L$.
Step 2: Key Formula or Approach:
We apply Kirchhoff's Voltage Law (KVL) along the circuit segment from terminal $A$ to terminal $B$.
As we move in the direction of the current $I$:
1. The potential drop across the internal resistance is given by Ohm's Law: $V_R = IR$.
2. The induced back-emf dropping across the inductor coil is given by Faraday-Lenz laws: $V_L = L\frac{dI}{dt}$.
The total loop branch potential equation is:
$$V_A - IR - L\frac{dI}{dt} = V_B \implies V_A - V_B = L\frac{dI}{dt} + IR$$
Step 3: Detailed Explanation:
Let's substitute the given numerical parameters into our branch potential equation:
$$0.5 = L(8) + (0.5 \times 0.2)$$
Calculate the resistive voltage drop term ($IR$):
$$0.5 \times 0.2 = 0.1\text{ V}$$
Substitute this value back into the equation:
$$0.5 = 8L + 0.1$$
Isolate the term containing our unknown inductor parameter, $8L$:
$$8L = 0.5 - 0.1$$
$$8L = 0.4$$
Divide both sides by 8 to solve for $L$:
$$L = \frac{0.4}{8} = \frac{4}{80} = 0.05\text{ H}$$
Step 4: Final Answer:
The self-inductance of the coil is $0.05\text{ H}$, which matches option (D).
