two cells of emf's 4.5v and 6v and resistance 6 & 3 Ohm have their -ve terminals joined by a wire of 18 Ohm and +ve terminal joines by a wire f 12 ohm . A third resisitance of wire 24 ohms of these wires . Draw circuit diagram and using kirchoffs rule find potential difference at the end of the third wire

Refer to the following circuit diagram

we have to find the potential difference across RS.

now here

by applying Kirchoff's Law in loop PQRS, we have

6I₁ + 6I₁ + 24(I₁ + I₂) + 9I₁ = 4.5  .......................(1)

so,

45I₁ + 24I₂ = 4.5

or dividing both sides by 3, we get

15I₁ + 8I₂ = 1.5  ............................(2)

and

by applying Kirchoff's Law in loop RSNM, we have

24(I₁ + I₂) + 9I₂ + 3I₂ + 6I₂ = 6  ................................(3)

so,

24I₁ + 42I₂ = 6

or dividing both sides by 3, we get

8I₁ + 14I₂ = 2  ............................(4)

now, we can solves for I₁ and I₂ so, by doing that we get

I₁ = 0.034 amps

I₂ = 0.123 amps

so,

the potential difference across RS will be

V_RS = (I₁ + I₂) x 24ohms = (0.034 + 0.123) x 24

thus,

V_RS = 3.768 volts