In the given circuit,if the potential at point $B$ is $24 \ V$,the potential at point $A$ is: (in $V$)

In the given figure,the value of $V_{0}$ will be $.....V$.

The resistances of the four arms $P, Q, R$ and $S$ in a Wheatstone's bridge are $10 \, \Omega$, $30 \, \Omega$, $30 \, \Omega$ and $90 \, \Omega$, respectively. The e.m.f. and internal resistance of the cell are $7 \, V$ and $5 \, \Omega$ respectively. If the galvanometer resistance is $50 \, \Omega$, the current drawn from the cell will be ............... $A$.

In the network shown,the cell of emf $E$ has internal resistance $r$ and the galvanometer shows zero deflection. If the cell is replaced by a new cell of emf $2E$ and internal resistance $3r$,keeping everything else identical,then:

If a current of $7 \ A$ flows through the electric circuit,what is the potential difference between $B$ and $C$ in volts?

Two ideal batteries of emf $V_1$ and $V_2$ and three resistances $R_1, R_2$ and $R_3$ are connected as shown in the figure. The current in resistance $R_2$ would be zero if:
$(A)$ $V_1=V_2$ and $R_1=R_2=R_3$
$(B)$ $V_1=V_2$ and $R_1=2R_2=R_3$
$(C)$ $V_1=2V_2$ and $2R_1=2R_2=R_3$
$(D)$ $2V_1=V_2$ and $2R_1=R_2=R_3$