Q153 / 180 · KCET 2011 · Physics
In the circuit shown, the currents $i_{1}$ and $i_{2}$ are
- A$i_{1}=1.5 \mathrm{~A}, i_{2}=0.5 \mathrm{~A}$
- Correct answerB$i_{1}=0.5 \mathrm{~A}, i_{2}=1.5 \mathrm{~A}$
- C$i_{1}=1 \mathrm{~A}, i_{2}=3 \mathrm{~A}$
- D$i_{1}=3 \mathrm{~A}, i_{2}=1 \mathrm{~A}$
Solution
Resistance $R=\frac{12 \times 4}{12+4}+2=\frac{48}{16}+2=5 \Omega$ Electric current $I=\frac{E}{R+r}=\frac{12}{6}=2 \mathrm{~A}$
$$
\begin{gathered}
I_{1}+I_{2}=2 \mathrm{~A} \\
I \propto \frac{1}{R} \\
\therefore \quad I_{2}=1.5 \mathrm{~A}, I_{1}=0.5 \mathrm{~A}
\end{gathered}
$$
Hence, the correct answer is option: 'B'.