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In the circuit shown in the figure, the current ' $T$ ' is
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Verified Answer
The correct answer is:
$6 \mathrm{~A}$
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
$$
\begin{gathered}
\Delta V=I R \\
\Delta V=24-5.14=18.86, R=3 \Omega \\
\therefore \quad I=\frac{18.86}{3} \approx 6 \mathrm{~A}
\end{gathered}
$$
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
$$
\begin{gathered}
\Delta V=I R \\
\Delta V=24-5.14=18.86, R=3 \Omega \\
\therefore \quad I=\frac{18.86}{3} \approx 6 \mathrm{~A}
\end{gathered}
$$
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