AP EAMCET · PHYSICS · Current Electricity
A \(6 \mathrm{~V}\) cell with \(0.5 \Omega\) internal resistance, a \(10 \mathrm{~V}\) cell with \(1 \Omega\) internal resistance and a \(12 \Omega\) external resistance are connected in parallel. The current (in ampere) through the \(10 \mathrm{~V}\) cell is
- A 0.6
- B 2.27
- C 2.87
- D 5.14
Answer & Solution
Correct Answer
(C) 2.87
Step-by-step Solution
Detailed explanation
In closed loop \(A B G F E H A\), \(\begin{aligned} 10-i_2 \times 1+i_1 \times 0.5-6 & =0 \\ 0.5 i_1-i_2 & =-4 \end{aligned}\) In closed loop \(B C D E B\), \(\begin{array}{r} \left(i_1+i_2\right) \times 12+i_2 \times 1-10=0 \\ 12 i_1+13 i_2=10 \end{array}\) From Eqs. (i) and…
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