JEE Mains 2020 Model Paper 4 New Pattern

The potential differenced across 5 µC capacitor, is V =

Q

C

= (

1000

5

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

V

R

=

200]

10

= 20 A
From KCL at junction A we get I = 25 A

Now we apply KVL in the loop ABDA , - (10) (20) - 5R2 + (50) (5) = 0 ⇒ R2 = 10 Ω
Apply KVL in loop ADCEFA, we get - (50) (5) - (5) (10) + 310 - 25R1 = 0
⇒ R1 =

10

25

Ω = 0.4 Ω
Apply KVL in loop ABCDA,
- (20) (10) - 15 R3 + (10) (5) + (50) (5) = 0
⇒ R3 =

100

15

= 6.66 Ω