In the circuit shown,the cells $A$ and $B$ have negligible resistance. For $V_{A} = 12 \; V$,$R_{1} = 500 \; \Omega$ and $R = 100 \; \Omega$,the galvanometer $(G)$ shows no deflection. The value of $V_{B}$ is .... $V$.

$A$ voltmeter is connected in parallel with a variable resistance $R$ which is in series with an ammeter and a cell as shown in the figure. For one value of $R$,the meters read $0.3 \, A$ and $0.9 \, V$. For another value of $R$,the readings are $0.25 \, A$ and $1.0 \, V$. What is the internal resistance of the cell in $\Omega$?

Refer to the circuit shown. What will be the total power dissipation in the circuit if $P$ is the power dissipated in $R_1$ (in $P$)? It is given that $R_2 = 4 R_1$ and $R_3 = 12 R_1$.

In the circuit shown in the figure,switch $S_1$ is initially closed and $S_2$ is open. Find $V_a - V_b$ in $V$.

Four identical electrical lamps are labelled $1.5\,V$,$0.5\,A$,which describes the condition necessary for them to operate at normal brightness. $A$ $12\,V$ battery of negligible internal resistance is connected to the lamps as shown. Then:

$A$ $16 \ \Omega$ wire is bent to form a square loop. $A$ $9 \ V$ battery with internal resistance $1 \ \Omega$ is connected across one of its sides. If a $4 \ \mu F$ capacitor is connected across one of its diagonals,the energy stored by the capacitor will be $\frac{x}{2} \ \mu J$,where $x = . . . . . . .$