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 ________.
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Use Kirchhoff's Voltage Law: the voltage across the inductor is the battery EMF minus the voltage drop across its resistance ($V_L = E - iR$).
An inductor with resistance acts as a series combination of a pure inductor ($L$) and a resistor ($R$). When the circuit is closed, the rate of change of current determines the voltage across the inductor portion. The governing equation for the circuit is: $$ L \frac{di}{dt} + iR = E $$ The voltage across the inductor is specifically the induced EMF, which is $V_L = L \frac{di}{dt}$. From the circuit equation, we can see that: $$ V_L = E - iR $$
Step 1: Calculate the voltage for the first current value. For $i = 2 \text{ mA} = 0.002 \text{ A}$: $$ V_1 = 1 - (0.002)(100) = 1 - 0.2 = 0.8 \text{ V} $$
Step 2: Calculate the voltage for the second current value. For $i = 4 \text{ mA} = 0.004 \text{ A}$: $$ V_2 = 1 - (0.004)(100) = 1 - 0.4 = 0.6 \text{ V} $$
Step 3: Find the ratio between the two calculated voltages. $$ \frac{V_1}{V_2} = \frac{0.8}{0.6} $$ By simplifying the fraction by dividing both numerator and denominator by $0.2$, we get: $$ \frac{V_1}{V_2} = \frac{4}{3} $$ The ratio is $4/3$, matching option 1.
