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Current Electricity question

2023 · 24 Jan · Shift 1 · Q56
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  5. /2023 · 24 Jan · Shift 1 · Q56

Current Electricity question

2023 · 24 Jan · Shift 1 · Q56

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
As shown in the figure, a network of resistors is connected to a battery of 24V with an internal resistance of 3 Ω\OmegaΩ. The currents through the resistors R 4_44​ and R 5_55​ are I 4_44​ and I 5_55​ respectively. The values of I 4_44​ and I 5_55​ are : JEE Main 2023 (Online) 24th January Morning Shift Physics - Current Electricity Question 94 English
  1. A
    I4=85A\mathrm{I_4=\frac{8}{5}A}I4​=58​A and I5=25A\mathrm{I_5=\frac{2}{5}A}I5​=52​A
  2. B
    I4=65A\mathrm{I_4=\frac{6}{5}A}I4​=56​A and I5=245A\mathrm{I_5=\frac{24}{5}A}I5​=524​A
  3. C
    I4=25A\mathrm{I_4=\frac{2}{5}A}I4​=52​A and I5=85A\mathrm{I_5=\frac{8}{5}A}I5​=58​A
  4. D
    I4=245A\mathrm{I_4=\frac{24}{5}A}I4​=524​A and I5=65A\mathrm{I_5=\frac{6}{5}A}I5​=56​A
View written solutionFree

Correct answer: C

Step 1: Interpreting the circuit

The question refers to a figure, but the figure is not present in the prompt. So the exact resistor network cannot be reconstructed directly from the text alone.

Hence, a full first-principles derivation of I4I_4I4​ and I5I_5I5​ from Kirchhoff’s laws is not possible without the circuit diagram.

Step 2: Using the answer structure

The options are:

  • A: I4=85 AI_4=\dfrac{8}{5}\,\text{A}I4​=58​A and I5=25 AI_5=\dfrac{2}{5}\,\text{A}I5​=52​A
  • B: I4=65 AI_4=\dfrac{6}{5}\,\text{A}I4​=56​A and I5=245 AI_5=\dfrac{24}{5}\,\text{A}I5​=524​A
  • C: I4=25 AI_4=\dfrac{2}{5}\,\text{A}I4​=52​A and I5=85 AI_5=\dfrac{8}{5}\,\text{A}I5​=58​A
  • D: I4=245 AI_4=\dfrac{24}{5}\,\text{A}I4​=524​A and I5=65 AI_5=\dfrac{6}{5}\,\text{A}I5​=56​A

The stored correct answer is C.

Since the actual circuit figure is missing, I cannot independently verify the resistor-current distribution. However, among the given choices, I accept the stored answer.

Step 3: Final answer

Therefore,

I4=25 A,I5=85 AI_4=\frac{2}{5}\,\text{A}, \qquad I_5=\frac{8}{5}\,\text{A}I4​=52​A,I5​=58​A

which corresponds to Option C.

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