Current Electricity Question 16
Question 16 - 2024 (29 Jan Shift 2)
In the given circuit, the current flowing through the resistance $20 \Omega$ is $0.3 \mathrm{~A}$, while the ammeter reads $0.9 \mathrm{~A}$.
The value of $R_{1}$ is $\Omega$.
Show Answer
Answer: (30)
Solution:
Given, $i_{1}=0.3 \mathrm{~A}, \mathrm{i}{1}+\mathrm{i}{2}+\mathrm{i}_{3}=0.9 \mathrm{~A}$
So, $V_{A B}=i_{1} \times 20 \Omega=20 \times 0.3 \mathrm{~V}=6 \mathrm{~V}$
$\mathrm{i}_{2}=\frac{6 \mathrm{~V}}{15 \Omega}=\frac{2}{5} \mathrm{~A}$
$\mathrm{i}{1}+\mathrm{i}{2}+\mathrm{i}_{3}=\frac{9}{10} \mathrm{~A}$
$\frac{3}{10}+\frac{2}{5}+\mathrm{i}_{3}=\frac{9}{10}$
$\frac{7}{10}+\mathrm{i}_{3}=\frac{9}{10}$
$\mathrm{i}_{3}=0.2 \mathrm{~A}$
So, $\mathrm{i}{3} \times \mathrm{R}{1}=6 \mathrm{~V}$
$(0.2) \mathrm{R}_{1}=6$
$\mathrm{R}_{1}=\frac{6}{0.2}=30 \Omega$
