In a Wheatstone network,$P=2 \Omega, Q=2 \Omega, R=2 \Omega$,and $S=3 \Omega$. The resistance with which $S$ is to be shunted so that the bridge may be balanced is: (in $\Omega$)

Five equal resistances each of $2R$ are connected as shown in the figure. $A$ battery of $V$ volts is connected between $A$ and $B$. Then,the current through $FC$ is:

In the given figure: $V_1=V, V_2=\alpha V, R_1=\beta R, R_2=\gamma R$,where $\alpha, \beta$,and $\gamma$ are positive real numbers. The value of current $I$ is

The potential difference $(V_A - V_B)$ between the points $A$ and $B$ in the given figure is (in $V$)

Consider the circuit shown in the figure where all the resistances are of magnitude $1 k \Omega$. If the current in the extreme right resistance $X$ is $1 mA$, the potential difference between $A$ and $B$ is: (in $V$)

In the circuit shown in the figure,the current $I$ is (in $\,A$)