Q77 / 90 · JEE Main 2024 (29 Jan Shift 2) · Physics

In the given circuit, the current flowing through the resistance $20 Ω$ is $0.3 A$ , while the ammeter reads $0.9 A$ . The value of $R_{1}$ is _____ $Ω$ .

Correct answer: 30

Solution

Given, $i_{1} = 0.3 A, i_{1} + i_{2} + i_{3} = 0.9 A$

So, $V_{AB} = i_{1} \times 20 Ω = 0.3 \times 20 V = 6 V$

$i_{2} = \frac{6 V}{15 Ω} = \frac{2}{5} A$

Now,

$i_{1} + i_{2} + i_{3} = \frac{9}{10} A$

$\Rightarrow \frac{3}{10} + \frac{2}{5} + i_{3} = \frac{9}{10}$

$\Rightarrow \frac{7}{10} + i_{3} = \frac{9}{10}$

$\Rightarrow i_{3} = 0.2 A$

So, $i_{3} \times R_{1} = 6 V$

$\Rightarrow (0.2) R_{1} = 6$

$\Rightarrow R_{1} = \frac{6}{0.2} = 30 Ω$

Hence, the correct answer is 30.