JEE Mains · Physics · STD 12 - 3. current electricity
As shown in the figure, a network of resistors is connected to a battery of \(2\,V\) with an internal resistance of \(3\,\Omega\). The currents through the resistors \(R_4\) and \(R_5\) are \(I_4\) and \(I_5\) respectively. The values of \(I_4\) and \(I_5\) are :
- A \(I _4=\frac{8}{5} A\) and \(I _5=\frac{2}{5}\,A\)
- B \(I _4=\frac{24}{5} A\) and \(I _5=\frac{6}{5}\,A\)
- C \(I _4=\frac{6}{5} A\) and \(I _5=\frac{24}{5}\,A\)
- D \(I _4=\frac{2}{5} A\) and \(I _5=\frac{8}{5}\,A\)
Answer & Solution
Correct Answer
(D) \(I _4=\frac{2}{5} A\) and \(I _5=\frac{8}{5}\,A\)
Step-by-step Solution
Detailed explanation
Equivalent resistance of circuit \(R _{ eq } =3+1+2+4+2\) \(=12 \Omega\) Current through battery \(i =\frac{24}{12}=2 A\) \(I _4=\frac{ R _5}{ R _4+ R _5} \times 2=\frac{5}{20+5} \times 2=\frac{2}{5} A\) \(I _5=2-\frac{2}{5}=\frac{8}{5} A\)
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