AP EAMCET · PHYSICS · Current Electricity
In the circuit, \(E_1=E_2=E_3=2 V\) and \(R_1\) and \(\mathrm{R}_2=4 \mathrm{~W}\). Then the current flowing through \(\mathrm{E}_2\) is
- A Zero
- B A from \(A\) to \(B\)
- C \(4 \mathrm{~A}\) from \(\mathrm{A}\) to \(\mathrm{B}\)
- D \(2 \mathrm{~A}\) from \(\mathrm{B}\) to \(\mathrm{A}\)
Answer & Solution
Correct Answer
(D) \(2 \mathrm{~A}\) from \(\mathrm{B}\) to \(\mathrm{A}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & E_{e q}=\frac{\frac{E_1}{R_1}+\frac{E_3}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}}=\frac{E_1 R_2+E_3 R_1}{R_1+R_2}=\frac{2 \times 4+2 \times 4}{42+4}=2 V \\ & R_{e q}=\frac{R_1 R_2}{R_1+R_2}=\frac{4 \times 4}{4 \times 4}=2 \Omega \\ & -E_2+i R_{e q}-E_{e q}=0 \\ &…
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