Physics - Current Electricity Question with Solution | TestHub

An equivalent of the given network is as shown in the figure. If $R_p$ be the net resistance, then
$\begin{array}{rl}\frac{1}{{\mathrm{R}}_{\mathrm{p}}}& =\frac{1}{10}+\left(\frac{1}{10+10}\right)+\left(\frac{1}{10+10}\right)+\frac{1}{10}\\ & =\frac{1}{10}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}\end{array}$ $\begin{array}{l}=\frac{6}{20}=\frac{3}{10}\\ \therefore {\mathrm{R}}_{\mathrm{p}}=\frac{10}{3}\Omega \end{array}$ Hence, current flowing through ammeter is $$I=\frac{V}{R_p}=\frac{12}{\left(\frac{10}{3}\right)}=3.6\mathrm{A}$$