Question 149: AC fundamentals

Question

In the above circuit, C = ${{\sqrt 3 } \over 2}$$\mu $F, R
2
= 20 $\Omega $, L = ${{\sqrt 3 } \over {10}}$ H and R
1
= 10 $\Omega $. Current in L-R
1
path is I
1
and in C-R
2
path it is I
2
. The voltage of A.C. source is given by, V = 200 ${\sqrt 2 }$ sin (100 t) volts . The phase difference between I
1
and I
2
is :
![](https://imagex.cdn.examgoal.net/yqHnRVwUAcGd3oP64/atYXMsjq2p6JEN4dqZKecQBgdUTdZ/zMAPFW6dafK7nhmIPxIHQ5/image.png)

Accepted Answer

Phase difference between I
2
and V, i.e. C $-$ R
2
circuit is given by
$	an \phi  = {{{X_C}} \over {{R_2}}} \Rightarrow 	an \phi  = {1 \over {C\omega {R_2}}}$
Substituting the given values, we get
$	an \phi  = {1 \over {{{\sqrt 3 } \over 2} 	imes {{10}^{ - 6}} 	imes 100 	imes 20}} = {{{{10}^3}} \over {\sqrt 3 }}$
$	herefore$ $\phi$
1
is nearly 90$^\circ$.
Phase difference between I
1
and V, i.e. in L $-$ R
1
circuit is given by
$	an {\phi _2} =  - {{{X_L}} \over {{R_1}}} =  - {{L\omega } \over R}$
Substituting the given values, we get
$	an {\phi _2} =  - {{{{\sqrt 3 } \over {10}} 	imes 100} \over {10}} =  - \sqrt 3 $
As, $	an {\phi _2} =  - \sqrt 3 $
$	herefore$ ${\phi _2} = 120^\circ $
Now, phase difference between I
1
and I
2
is
$\Delta \phi  = {\phi _2} - {\phi _1} = 120^\circ  - 90^\circ  = 30^\circ $

Answer

150
o

Answer

90
o

Answer

30
o

Answer

0
o

AC fundamentals

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