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  • A constant voltage of $50 V$ is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is

Q.7) A constant voltage of $50 V$ is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :

A) 3.0 A

B) 1.5 A

C) 2.0 A

D) 2.5 A

 

Answers (1)

best_answer

Correct Option: (1) $3.0 A$
Solution:
Let the current in the top loop be $\mathbf{I}_{\mathbf{1}}$ and in the bottom loop be $\mathbf{I}_{\mathbf{2}}$, both clockwise.
Then current through branch $C D(3 \Omega)=I_1-I_2$.

$$
\begin{aligned}
& \text { Loop } 1 (A \rightarrow 1 \Omega \rightarrow C \rightarrow 2 \Omega \rightarrow B \rightarrow A) \text { : } \\
& 1 \times I_1+3 \times\left(l_1-I_2\right)+2 \times\left.\right|_ 1=50 \\
& \Rightarrow I_1(1+3+2)-3 I_2=50 \\
& \Rightarrow 6 I_1-3 I_2=50 \ldots(1) \\ 
& \text { Loop } 2(A \rightarrow 3 \Omega \rightarrow D \rightarrow 4 \Omega \rightarrow B \rightarrow A) \text { : } \\
& 3 \times I_2+3 \times\left(I_2-I_1\right)+4 \times I_2=50 \\
& \Rightarrow I_2(3+3+4)-3 I_1=50 \\
& \Rightarrow 10 I_2-3 I_1=50 \ldots(2)
\end{aligned}
$$


Now, multiply equation (1) by 10 and (2) by 3 :
(1) $\times 10: 60 \mathrm{l}_1-30 \mathrm{I}_2=500$
(2) $\times 3:-9 I_1+30 I_2=150$

Add the two equations:

$$
\begin{aligned}
& \left(60 l_1-9 l_1\right)+\left(-30 l_2+30 l_2\right)=500+150 \\
& 51 l_1=650 \\
& l_1=650 / 51 \approx 12.75 \mathrm{~A}
\end{aligned}
$$


Now from equation (1):

$$
\begin{aligned}
& 6(12.75)-31_2=50 \\
& 76.5-31_2=50 \\
& 31_2=26.5 \\
& \mathrm{I}_2=26.5 / 3 \approx 8.83 \mathrm{~A}
\end{aligned}
$$


Current through $C D=I_1-I_2=12.75-8.83 \approx 3 \mathrm{~A}$
Hence, the answer is option (1) 3.0 A.

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Dimpy

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