SATHEE: Electromagnetic Induction and Alternating Current 4 Question 19

Electromagnetic Induction and Alternating Current 4 Question 19

####19. An inductor of inductance $L = 400 m H$ and resistors of resistances $R_{1} = 2 Ω$ and $R_{2} = 2 Ω$ are connected to a battery of emf $E = 12 V$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at time $t = 0$ .

What is the potential drop across $L$ as a function of time? After the steady state is reached, the switch is opened. What is the direction and the magnitude of current through $R_{1}$ as a function of time?

$(2001, 5 M)$

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Answer:

Correct Answer: 19. $12 e^{- 5 t} V, 6 e^{- 10 t} A (clockwise)$

Solution:

(a) Given, $R_{1} = R_{2} = 2 Ω, E = 12 V$

and $L = 400 m H = 0.4 H$ . Two parts of the circuit are in parallel with the applied battery. So, the upper circuit can be broken as

(a)

(b)

Now, refer Fig. (b)

This is a simple $L - R$ circuit, whose time constant

$τ_{L} = L / R_{2} = \frac{0.4}{2} = 0.2 s$

and steady state current

$i_{0} = \frac{E}{R_{2}} = \frac{12}{2} = 6 A$

Therefore, if switch $S$ is closed at time $t = 0$ , then current in the circuit at any time $t$ will be given by

$\begin{aligned} i (t) & = i_{0} (1 - e^{- t / τ_{L}}) \\ i (t) & = 6 (1 - e^{- t / 0.2}) \\ = 6 (1 - e^{- 5 t}) = i \end{aligned}$

Therefore, potential drop across $L$ at any time $t$ is

$\begin{aligned} V = L \frac{d i}{d t} = L (30 e^{- 5 t}) = (0.4) (30) e^{- 5 t} \\ V = 12 e^{- 5 t} V \end{aligned}$

(b) The steady state current in $L$ or $R_{2}$ is

$i_{0} = 6 A$

Now, as soon as the switch is opened, current in $R_{1}$ is reduced to zero immediately. But in $L$ and $R_{2}$ it decreases exponentially. The situation is as follows