Electromagnetic Induction and Alternating Current 4 Question 22
####22. A circuit containing a two position switch $S$ is shown in figure.
(1991, 4+4M)
(a) The switch $S$ is in position 1. Find the potential difference $V _A-V _B$ and the rate of production of joule heat in $R _1$.
(b) If now the switch $S$ is put in position 2 at $t=0$. Find
(i) steady current in $R _4$ and
(ii) the time when current in $R _4$ is half the steady value. Also, calculate the energy stored in the inductor $L$ at that time.
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Answer:
Correct Answer: 22. $(a) $-5V , 24.5W $
(b) (i) $0.6A $
(ii) $1.386 \times {10} ^{-3} s$, $4.5 \times {10} ^{-4} J$
Solution:
- (a) In steady state, no current will flow through capacitor. Applying Kirchhoff’s second law in loop 1
$$ \begin{aligned} & -2 i _2+2\left(i _1-i _2\right)+12=0 \\ & \therefore \quad 2 i _1-4 i _2=-12 \\ & \text { or } \quad i _1-2 i _2=-6 \end{aligned} $$
Applying Kirchhoff’s second law in loop 2
$$ -12-2\left(i _1-i _2\right)+3-2 i _1=0 $$
$$ \therefore \quad 4 i _1-2 i _2=-9 $$
Solving Eqs. (i) and (ii), we get
$$ \begin{aligned} i _2 & =2.5 A \text { and } i _1=-1 A \\ \text { Now, } V _A+3-2 i _1 & =V _B \\ \text { or } V _A-V _B=2 i _1-3 & =2(-1)-3=-5 V \\ P _{R _1}=\left(i _1-i _2\right)^{2} R _1 & =(-1-2.5)^{2}(2)=24.5 W \end{aligned} $$
(b) In position 2 : Circuit is as under
(i) Steady current in $R _4$ :
$$ i _0=\frac{3}{3+2}=0.6 A $$
(ii) Time when current in $R _4$ is half the steady value
$$ \begin{aligned} t _{1 / 2} & =\tau _L(\ln 2)=\frac{L}{R} \ln (2) \\ & =\frac{\left(10 \times 10^{-3}\right)}{5} \ln (2) \\ & =1.386 \times 10^{-3} s \\ U & =\frac{1}{2} L i^{2}=\frac{1}{2}\left(10 \times 10^{-3}\right)(0.3)^{2} \\ & =4.5 \times 10^{-4} J \end{aligned} $$
