Question 31: Electromagnetic Induction

Question

An inductor of inductance 10 mH having resistance of $100 \Omega$ is connected to battery of E.M.F. 1.0 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 mA and 4 mA is $\_\_\_\_$ .

Accepted Answer

When a battery is connected to an inductor $(\mathrm{L})$ and a resistor $(\mathrm{R})$ in series, the total voltage $(\mathrm{E})$ provided by the battery is shared between the resistance and the inductance according to Kirchhoff's Voltage Law:
$$ E=V_R+V_L $$
Where, E is the E.M.F. of the battery. $\mathrm{V}_{\mathrm{R}}$ is the voltage across the resistance, calculated as $\mathrm{V}_{\mathrm{R}}=\mathrm{iR} . \mathrm{V}_{\mathrm{L}}$ is the instantaneous voltage across the inductor.
$$ V_L=E-i R $$
The emf of the battery is $\mathrm{E}=1.0 \mathrm{~V}$ and connected resistance is $\mathrm{R}=100 \Omega$.
Case 1 when the current passing through inductor is $\mathrm{i}_1=2 \mathrm{~mA}=2 	imes 10^{-3} \mathrm{~A}$.
$$ V_{L 1}=1.0-\left(2 	imes 10^{-3} 	imes 100ight)=1.0-0.2=0.8 \mathrm{~V} $$
Case 2 when the current passing through inductor is $\mathrm{i}_2=4 \mathrm{~mA}=4 	imes 10^{-3} \mathrm{~A}$.
$$ V_{L 2}=1.0-\left(4 	imes 10^{-3} 	imes 100ight)=1.0-0.4=0.6 \mathrm{~V} $$
So, the ratio of the instantaneous voltages is:
$$ \frac{V_{L 1}}{V_{L 2}}=\frac{0.8}{0.6}=\frac{8}{6}=\frac{4}{3} $$
Therefore, the correct option is (A).

Answer

4/3

Answer

3/4

Answer

5/3

Answer

3/5

Electromagnetic Induction

An inductor of inductance 10 mH having resistance of 100 \Omega is connected to battery of E.M.F. 1.0 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 mA and 4 mA is \_\_\_\_ .

Choose the correct answer: