A circuit containing a two position switch S is shown in fig.

(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 stroed in the inductor L at that time.

Question

A circuit containing a two position switch S is shown in fig.

(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 stroed in the inductor L at that time.

Filo · Accepted Answer

Answer: A::B::C::DSolution: (a) (i) In this case S and I are connected
Solution Image

Using Kirchhoff's law in ABCDGA

+ 3 - (I_{2}) \times 2 - 12 + (I_{1}) \times 2 = 0

2 (I_{1}) - 2 (I_{2}) = 9

...(i)
Applying Kirchhoff's law in DEFGD

= 2 (I_{1}) + 12 - (I_{1} + I_{2}) 2 = 0

...(ii)
From (i) and (ii)

(I_{1}) = \frac{21}{6} a m p . F r o m (i i)

(I_2)=-1amp

. T o f \in d p o t e n t i a l d \Leftrightarrow e r e n c e b e t w e e n A and B

V_(A)+3-(-1)xx2=V_(B) implies (V_A)-(V_B)=-5V

T h e r a t e o f \prod u c t i o n o f h e a t o f h e a t \in

(R_1)

=(I_(1)^(2))(R_1)=(21/6)^(2)xx2=24.5w

(i) w h e n t h e s w i t c h i s p u t \in p o s i t i o n 2 t h e n t h e a c t i v e \circ u i t w i l l b e a s s h o w n \in t h e f i g u r e . W h e n t h e s t e a d y s t a t e c u r r e n t i s r e a c h e d t h e n t h e \in d u c \to r p l a y s n o r o \leq\in t h e \circ u i t .

(E_2)=I(R_(2)+R_(4))

implies I=3/5=0.6amp.

(i i) t h e g r o w t h o f c u r r e n t \in L - R \circ u i t i s g i v e n b y t h e exp r e s s i o n

I=(I_0)[1-e^(-(R)/(L)t)]

W h e n

I=(I_0)/(2), then (I_0)/(2)=(I_0)[1-e^(-(R)/(L)t)]

\Rightarrow \frac{1}{2} = 1 - e^{- \frac{R}{L} t} \Rightarrow e^{- \frac{R}{L} t} = \frac{1}{2}

Taking log on both side

lo g_{e} e^{- \frac{R}{L} t} = \frac{lo g _{e} 1}{2}

When

R = (R_{2}) + (R_{4})

= t = 1.386 \times 10^{- 3} sec

.Thus this much time is required for current to reach hald of its steady value.
The energy stored by the inductor at that time is given
bu

E = \frac{1}{2} L I^{2} = \frac{1}{2} \times 10 \times 10^{- 3} \times (\frac{0.6}{2})^{2} = 4.5 \times 10^{- 4} J

.

Question Text	A circuit containing a two position switch S is shown in fig. (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 stroed in the inductor L at that time.
Answer Type	Text solution:1
Upvotes	152

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