Q150 / 160 · TS EAMCET 2013 · Physics
In the circuit shown in the figure, the current ' $T$ ' is
- Correct answerA$6 \mathrm{~A}$
- B$2 \mathrm{~A}$
- C$4 \mathrm{~A}$
- D$7 \mathrm{~A}$
Solution
Applying junction law
We have
$$
\begin{array}{rlrl}
I & =I_1+I_2 \\
\frac{24-V}{3} & =\frac{10-V}{2}+\frac{9-V}{1} \\
\Rightarrow & \frac{24-V}{3} & =\frac{28-3 V}{2} \\
\Rightarrow & 2(24-V) & =3(28-3 V) \\
\Rightarrow & 48-2 V & =84-9 \mathrm{~V} \\
\Rightarrow & 7 V & =36 \\
\Rightarrow & V & =5.14 \mathrm{~V}
\end{array}
$$
From Ohm's law
$$
\begin{gathered}
\Delta V=I R \\
\Delta V=24-5.14=18.86, R=3 \Omega \\
\therefore \quad I=\frac{18.86}{3} \approx 6 \mathrm{~A}
\end{gathered}
$$
Hence, the correct answer is option: 'A'.