In the given circuit,find the potential difference between $P$ and $Q$ in $V$.

The potential difference $V$ across the filament of the bulb shown in the given Wheatstone bridge varies as $V = i(2i + 1)$,where $i$ is the current in ampere through the filament of the bulb. The emf of the battery $(V_b)$ so that the bridge becomes balanced is (in $V$)

The heat generated in $1 \text{ minute}$ between points $A$ and $B$ in the given circuit, when a battery of $9 \text{ V}$ with internal resistance of $1 \ \Omega$ is connected across these points, is . . . . . . $J$.

The internal resistances of two cells shown are $0.1\,\Omega$ and $0.3\,\Omega$. If $R = 0.2\,\Omega$,the potential difference across the cell

In the given figure,when the galvanometer shows no deflection,the current (in $A$) flowing through the $5\,\Omega$ resistance will be:

In a balanced Wheatstone bridge,the current in the galvanometer is zero. It remains zero when: