Q.43. An inductor of inductance 10 mH having resistance of 100 \Omega is connected to battery of E.M

Q: What is the correct answer for JEE Main Physics Q43?

The correct answer is A. Given: Inductance $L = 10$ mH $= 10 \times 10^{-3}$ H Resistance $R = 100$ \Omega EMF of battery $E = 1.0$ V For an RL circuit, after the switch is closed, the current $i(t)$ at any time t is given by: $i(t) = \frac{E}{R} (1 - e^{-t/\tau})$, where $\tau = L/R$ is the time constant. The voltage across the inductor $V_L(t)$ is given by: $V_L(t) = L \frac{di}{dt}$ Alternatively, by KVL: $E = iR + V

Q: Q.43. 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…

The correct answer is A. Given: Inductance $L = 10$ mH $= 10 \times 10^{-3}$ H Resistance $R = 100$ \Omega EMF of battery $E = 1.0$ V For an RL circuit, after the switch is closed, the current $i(t)$ at any time t is given by: $i(t) = \frac{E}{R} (1 - e^{-t/\tau})$, where $\tau = L/R$ is the time constant. The voltage across the inductor $V_L(t)$ is given by: $V_L(t) = L \frac{di}{dt}$ Alternatively, by KVL: $E = iR + V

Q: What are the options for this JEE Main Physics question?

A. 4/3 · B. 3/4 · C. 5/3 · D. 3/5

Question

Q.43. 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 ____.

FirstInTest · Accepted Answer

Given:
Inductance $L = 10$ mH $= 10 	imes 10^{-3}$ H
Resistance $R = 100$ \Omega
EMF of battery $E = 1.0$ V

For an RL circuit, after the switch is closed, the current $i(t)$ at any time t is given by:
$i(t) = \frac{E}{R} (1 - e^{-t/	au})$, where $	au = L/R$ is the time constant.

The voltage across the inductor $V_L(t)$ is given by:
$V_L(t) = L \frac{di}{dt}$
Alternatively, by KVL: $E = iR + V_L$. So, $V_L = E - iR$.

We need to find the ratio of instantaneous voltages across the inductor for two different currents:
Case 1: Current $i_1 = 2$ mA $= 2 	imes 10^{-3}$ A
$V_{L1} = E - i_1 R = 1.0 	ext{ V} - (2 	imes 10^{-3} 	ext{ A}) 	imes (100 	ext{ \Omega})$
$V_{L1} = 1.0 - 0.2 = 0.8$ V.

Case 2: Current $i_2 = 4$ mA $= 4 	imes 10^{-3}$ A
$V_{L2} = E - i_2 R = 1.0 	ext{ V} - (4 	imes 10^{-3} 	ext{ A}) 	imes (100 	ext{ \Omega})$
$V_{L2} = 1.0 - 0.4 = 0.6$ V.

The ratio of instantaneous voltages across the inductor is:
$\frac{V_{L1}}{V_{L2}} = \frac{0.8 	ext{ V}}{0.6 	ext{ V}} = \frac{8}{6} = \frac{4}{3}$.

Answer

4/3

Answer

3/4

Answer

5/3

Answer

3/5

Related Questionsin Physics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions