Q124 / 125 · BITSAT 2020 · Physics
Determine the current in $2 \Omega$ resistor.
- A$1 \mathrm{~A}$
- B$1.5 \mathrm{~A}$
- Correct answerC$0.9 \mathrm{~A}$
- D$0.6 \mathrm{~A}$
Solution
At steady state the capacitor will be fully charged and thus there will be no current in the $1 \Omega$ resistance. So the effective circuit becomes
Net current from the $6 \mathrm{~V}$ battery, $I=\frac{6}{\left(\frac{2 \times 3}{2+3}\right)+\frac{2.8}{1}}=\frac{6}{1.2+2.8}=\frac{3}{2}=1.5 \mathrm{~A}$ Between $\mathrm{A}$ and $\mathrm{B}$, voltage is same in both resistances, $2 \mathrm{I}_{1}=3 \mathrm{I}_{2}$ where $\mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I}=1.5$ $\Rightarrow 2 \mathrm{I}_{1}=3\left(1.5-\mathrm{I}_{1}\right) \Rightarrow \mathrm{I}_{1}=0.9 \mathrm{~A}$
Net current from the $6 \mathrm{~V}$ battery, $I=\frac{6}{\left(\frac{2 \times 3}{2+3}\right)+\frac{2.8}{1}}=\frac{6}{1.2+2.8}=\frac{3}{2}=1.5 \mathrm{~A}$ Between $\mathrm{A}$ and $\mathrm{B}$, voltage is same in both resistances, $2 \mathrm{I}_{1}=3 \mathrm{I}_{2}$ where $\mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I}=1.5$ $\Rightarrow 2 \mathrm{I}_{1}=3\left(1.5-\mathrm{I}_{1}\right) \Rightarrow \mathrm{I}_{1}=0.9 \mathrm{~A}$
Hence, the correct answer is option: 'C'.