As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of $3 \Omega$. The currents through the resistors $\mathrm{R}_{4}$ and $\mathrm{R}_{5}$ are $\mathrm{I}_{4}$ and $\mathrm{I}_{5}$ respectively. The values of $\mathrm{I}_{4}$ and $\mathrm{I}_{5}$ are :

(D) $\mathrm{I}_{4}=\frac{2}{5} \mathrm{~A}$ and $\mathrm{I}_{5}=\frac{8}{5} \mathrm{~A}$

(A) $\mathrm{I}_{4}=\frac{8}{5} \mathrm{~A}$ and $\mathrm{I}_{5}=\frac{2}{5} \mathrm{~A}$

(B) $\mathrm{I}_{4}=\frac{24}{5} \mathrm{~A}$ and $\mathrm{I}_{5}=\frac{6}{5} \mathrm{~A}$

(C) $\mathrm{I}_{4}=\frac{6}{5} \mathrm{~A}$ and $\mathrm{I}_{5}=\frac{24}{5} \mathrm{~A}$

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Current Electricity

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

3Ω

. The currents through the resistors

R_{4}

and

R_{5}

are

I_{4}

and

I_{5}

respectively. The values of

I_{4}

and

I_{5}

are :

Answer: (D)

The two

2 Ω

resistors

R_{1}

and

R_{2}

in parallel give

1 Ω

, while

R_{4} = 20 Ω

with

R_{5} = 5 Ω

in parallel give

4 Ω

. Adding these to

R_{3} = 2 Ω

R_{6} = 2 Ω

and the internal

3 Ω

makes the total resistance

12 Ω

, so the battery drives

\frac{24}{12} = 2 A

. This current then splits between

R_{4}

and

R_{5}

in inverse proportion to their resistances, giving

I_{4} = 2 \times \frac{5}{25} = \frac{2}{5} A

and

I_{5} = 2 \times \frac{20}{25} = \frac{8}{5} A

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Single correct option