Question

An inductor of inductance $10 \text{ mH}$ having resistance of $100 \text{ } \Omega$ is connected to battery of E.M.F. $1.0 \text{ V}$ through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is $2 \text{ mA}$ and $4 \text{ mA}$ is ________.

• A. 4/3
• B. 3/4
• C. 5/3
• D. 3/5

Hint:

Use Kirchhoff's Voltage Law: the voltage across the inductor is the battery EMF minus the voltage drop across its resistance ($V_L = E - iR$).

Answer 1

The Correct Option is A

This problem involves the behavior of a real inductor in a DC circuit. A real inductor can be modeled as an ideal inductor with inductance $L$ in series with its internal resistance $R$. When a switch is closed in an $LR$ circuit connected to a battery of EMF $E$, Kirchhoff's Voltage Law (KVL) governs the instantaneous current and voltages.

According to KVL, the sum of the potential drops across the resistance and the inductor must equal the battery's EMF:
$$ E = iR + V_L $$
where $V_L$ is the instantaneous voltage across the inductive part of the inductor (i.e., $V_L = L \frac{di}{dt}$).

Rearranging this formula, the voltage across the inductor is:
$$ V_L = E - iR $$

We are given:
Battery EMF $E = 1.0 \text{ V}$
Internal resistance $R = 100 \text{ } \Omega$

Case 1: When the current $i_1 = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$
The voltage across the inductor is:
$$ V_{L1} = 1.0 - (2 \times 10^{-3} \times 100) $$
$$ V_{L1} = 1.0 - 0.2 = 0.8 \text{ V} $$

Case 2: When the current $i_2 = 4 \text{ mA} = 4 \times 10^{-3} \text{ A}$
The voltage across the inductor is:
$$ V_{L2} = 1.0 - (4 \times 10^{-3} \times 100) $$
$$ V_{L2} = 1.0 - 0.4 = 0.6 \text{ V} $$

The ratio of these instantaneous voltages is:
$$ \text{Ratio} = \frac{V_{L1}}{V_{L2}} = \frac{0.8}{0.6} = \frac{8}{6} = \frac{4}{3} $$

Thus, the ratio of the voltages is $4/3$.