Physics - Current Electricity Question with Solution | TestHub

$\mathrm{V}=\mathrm{IR}\Rightarrow 60=\mathrm{I}(20)\Rightarrow \mathrm{I}=3\mathrm{amp}$. When ${\mathrm{S}}_1\&{\mathrm{S}}_4$ are closed When ${\mathrm{S}}_2$ \& ${\mathrm{S}}_3$ are closed

By applying nodal analysis at junction x $\frac{\mathrm{x}-60}{10}+\frac{\mathrm{x}-60}{20}+\frac{\mathrm{x}-0}{10}=0$ $2\mathrm{x}-120+\mathrm{x}-60+2\mathrm{x}=0\Rightarrow \mathrm{x}=36$ $\mathrm{I}=\frac{60-36}{20}=\frac{6}{5}$

(C) When only S2 is closed, circuit will be balanced wheatstone bridge current in ammeter will be zero. (D) When all the switch are closed branch 'AB' is short circuited and current will become zero.