JEE MAIN - Physics (2024 - 9th April Morning Shift - No. 30)
The current flowing through the $$1 \Omega$$ resistor is $$\frac{n}{10}$$ A. The value of $$n$$ is _______.
Answer
25
Explanation
At $$C$$
$$\begin{aligned} & \frac{v_1}{2}+\frac{v_1-5}{2}+\frac{v_1+10-v_2}{1}=0 \\ & v_2=2 v_1+\frac{15}{2} \quad \text{.... (i)} \end{aligned}$$
At $$A$$
$$\begin{aligned} & \frac{v_2-5}{4}+\frac{v_2}{4}+\frac{v_2-10-v_1}{1}=0 \\ & 6 v_2=4 v_1+45 \quad \text{.... (ii)} \end{aligned}$$
$$\begin{aligned} & \Rightarrow v_1=0 \\ & \text { and } v_2=\frac{15}{2} \\ & \therefore \quad i=\frac{5}{2} \mathrm{~A} \\ & \Rightarrow n=25 \end{aligned}$$
Comments (0)
