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In given circuit, all resistances are of $10 \Omega$. Current flowing through ammeter is
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Verified Answer
The correct answer is:
$3.6 \mathrm{~A}$
An equivalent of the given network is as shown in the figure.
If $R_{p}$ be the net resistance, then
$\begin{aligned} \frac{1}{\mathrm{R}_{\mathrm{p}}} &=\frac{1}{10}+\left(\frac{1}{10+10}\right)+\left(\frac{1}{10+10}\right)+\frac{1}{10} \\ &=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10} \end{aligned}$
$\begin{array}{l}
\quad=\frac{6}{20}=\frac{3}{10} \\
\therefore \quad \mathrm{R}_{\mathrm{p}}=\frac{10}{3} \Omega
\end{array}$
Hence, current flowing through ammeter is
$$
I=\frac{V}{R_{p}}=\frac{12}{\left(\frac{10}{3}\right)}=3.6 \mathrm{~A}
$$
If $R_{p}$ be the net resistance, then
$\begin{aligned} \frac{1}{\mathrm{R}_{\mathrm{p}}} &=\frac{1}{10}+\left(\frac{1}{10+10}\right)+\left(\frac{1}{10+10}\right)+\frac{1}{10} \\ &=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10} \end{aligned}$
$\begin{array}{l}
\quad=\frac{6}{20}=\frac{3}{10} \\
\therefore \quad \mathrm{R}_{\mathrm{p}}=\frac{10}{3} \Omega
\end{array}$
Hence, current flowing through ammeter is
$$
I=\frac{V}{R_{p}}=\frac{12}{\left(\frac{10}{3}\right)}=3.6 \mathrm{~A}
$$
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