In the figure below, the switches S1 and S2 are closed simultaneously at...

2 months ago 1

In the figure below, the switches $S_{1} a n d S_{2}$ $S_{1} a n d S_{2}$ are closed simultaneously at $t = 0$ $t = 0$ and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current I in the middle wire reaches its maximum magnitude $I_{m a x}$ $I_{m a x}$ time $t = τ$ $t = τ$ . Which of the following statements is (are) true?

I_{m a x} = \frac{V}{2 R}

$I_{m a x} = \frac{V}{2 R}$

I_{m a x} = \frac{V}{4 R}

$I_{m a x} = \frac{V}{4 R}$

τ = \frac{L}{R} ln 2

$τ = \frac{L}{R} \ln 2$

τ = \frac{2 L}{R} l ln 2

$τ = \frac{2 L}{R} l \ln 2$

Solution:

The current in the middle, $I$ $I$ , as shown in the figure above has a maximum value,
$i_{m a x} = {(i_{2} - i_{1})}_{m a x}$ $i_{m a x} = {(i_{2} - i_{1})}_{m a x}$

Using the equation of current decay in an $L - R$ $L - R$ circuit, for the given values in the figure above,
$i = (i_{2} - i_{1}) = \frac{V}{R} [1 - e^{- (\frac{R}{2 L}) t}] - \frac{V}{R} [1 - e^{(- \frac{R}{L}) t}]$ $i = (i_{2} - i_{1}) = \frac{V}{R} [1 - e^{- (\frac{R}{2 L}) t}] - \frac{V}{R} [1 - e^{(- \frac{R}{L}) t}]$
$\frac{V}{R} [e^{(- \frac{R}{L}) t} - e^{(- \frac{R}{2 L}) t}]$ $\frac{V}{R} [e^{(- \frac{R}{L}) t} - e^{(- \frac{R}{2 L}) t}]$
For ${(i)}_{m a x}, \frac{d (i)}{d t} = 0$ ${(i)}_{m a x}, \frac{d (i)}{d t} = 0$
$\Rightarrow \frac{V}{R} [- \frac{R}{L} e^{- (\frac{R}{L}) t} - (- \frac{R}{2 L}) e^{- (\frac{R}{2 L}) t}] = 0$ $\Rightarrow \frac{V}{R} [- \frac{R}{L} e^{- (\frac{R}{L}) t} - (- \frac{R}{2 L}) e^{- (\frac{R}{2 L}) t}] = 0$
$\Rightarrow e^{- (\frac{R}{L}) t} = \frac{1}{2} e^{- (\frac{R}{2 L}) t}$ $\Rightarrow e^{- (\frac{R}{L}) t} = \frac{1}{2} e^{- (\frac{R}{2 L}) t}$
$\Rightarrow e^{- (\frac{R}{2 L}) t} = \frac{1}{2}$
$\Rightarrow (\frac{R}{2 L}) t = l n 2$
$t = \frac{2 L}{R} l n 2 \to$ time when I is maximum
$i_{m a x} = \frac{V}{R} [e^{- \frac{R}{L} (\frac{2 L}{R} l n 2)} - e^{- (\frac{R}{2 L}) (\frac{2 L}{R} l n 2)}]$
$\Rightarrow |i_{m a x}| = \frac{V}{R} |[\frac{1}{4} - \frac{1}{2}]| = \frac{1}{4} \frac{V}{R}$

Score

Ranking

Find Colleges

Compare Colleges

Exams

College Predictor

Course Finder

Your Gateway to Top Colleges & Exams

IIT Madras

IIT Delhi

IIT Bombay

IIT Kanpur

IIT Kharagpur

IIT Roorkee

IIT Guwahati

IIT Hyderabad

IIT BHU

IIT Dhanbad

IIT Indore

IIT Gandhinagar

IIT Ropar

IIT Jodhpur

IIT Mandi

IIT Patna

Top Exams

In the figure below, the switches S1 and S2 are closed simultaneously at...

Ranking

Find Colleges

Compare Colleges

Exams

College Predictor

Course Finder

Your Gateway to Top Colleges & Exams

IIT Madras

IIT Delhi

IIT Bombay

IIT Kanpur

IIT Kharagpur

IIT Roorkee

IIT Guwahati

IIT Hyderabad

IIT BHU

IIT Dhanbad

IIT Indore

IIT Gandhinagar

IIT Ropar

IIT Jodhpur

IIT Mandi

IIT Patna

Top Exams

In the figure below, the switches S1 and S2 are closed simultaneously at...

Join the conversation