As shown in the figure, a network of resistors is connected to a battery of 24   V with an internal…

As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of 3 Ω. The currents through the resistors R4 and R5 are I4 and I5 respectively. The values of I4 and I5 are:

  1. I4=85 A and I5=25 A
  2. I4=245 A and I5=65 A
  3. I4=65 A and I5=245 A
  4. I4=25 A and I5=85 A

Solution

Equivalent resistance of circuit
Req=3+2×22+2+2+20×520+5+2=12 Ω

Current through battery i=2412=2 A

Now, I4=R5R4+R5×2=520+5×2=25 A

and I5=i-I4=2-25=85 A

Asked in: JEE Main 2023 (24 Jan Shift 1)

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