AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

The figure below represents the given circuit.Vcap=

Q

C

Vcap=

1×10−3

5×10−6

Vcap=200V
The current across R5 resistance,
I=

V

R

I=

200

10

=20A
The figure below represents the current distribution.In loop BCDB,
−15R3+10×5+5R2=0 ......(I)
In loop ABCDA,
−200−15R3+50+250=0
15R3=100
R3=

100

15

Ω
From equation (I),
5R2=50
R2=10Ω
In loop ADCA,
−250−50+310−25R1=0
10=25R1
R1=

2

5

Ω
The required value is,

R1×R2

R3

=

2 3 ×10

( 100 15 )

=

3

5

=0.6