JEE Mains 2020 Model Paper 6 New Pattern

The potential differenced across 5µC capacitor, is V=

Q

C

=(

1000

5

)V=200V
The 10Ω resistor and capacitor are in parallel arrangement and therefore current I through resistor is given by I=

V

R

=

200

10

=20A
From KCL at junction A we get I=25ANow we apply KVL in the loop ABDA,
−10×20−5R2+50×5=0
⇒R2=10Ω
Apply KVL in loop ADCEFA, we get
−50×5−5×10+310−25R1=0
⇒R1=0.4Ω
Apply KVL in loop ABCDA,
−20×10−15R3+10×5+50×5=0
⇒R3=6.66Ω