Current Electricity Question 16
Question 16 - 2024 (29 Jan Shift 2)
In the given circuit, the current flowing through the resistance $20 \Omega$ is $0.3 A$, while the ammeter reads $0.9 A$.
The value of $R _1$ is $\Omega$.
Show Answer
Answer: (30)
Solution:
Given, $i _1=0.3 A, i _1+i _2+i _3=0.9 A$
So, $V _{A B}=i _1 \times 20 \Omega=20 \times 0.3 V=6 V$
$i _2=\frac{6 V}{15 \Omega}=\frac{2}{5} A$
$i _1+i _2+i _3=\frac{9}{10} A$
$\frac{3}{10}+\frac{2}{5}+i _3=\frac{9}{10}$
$\frac{7}{10}+i _3=\frac{9}{10}$
$i _3=0.2 A$
So, $i _3 \times R _1=6 V$
$(0.2) R _1=6$
$R _1=\frac{6}{0.2}=30 \Omega$
