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

Four resistors $A, B, C$ and $D$ form a Wheatstone bridge. The bridge is balanced when $C = 100 \ \Omega$. If $A$ and $B$ are interchanged,the bridge balances for $C = 121 \ \Omega$. The value of $D$ is: (in $\Omega$)

In a Wheatstone bridge,$P = 9 \, \Omega$,$Q = 11 \, \Omega$,$R = 4 \, \Omega$,and $S = 6 \, \Omega$. How much resistance $r$ must be connected in parallel to the resistance $S$ to balance the bridge?

For the circuit given below,the Kirchhoff's loop rule for the loop $BCDEB$ is given by the equation:

Resistances are arranged in a cyclic order to form a balanced Wheatstone bridge as shown in the figure. The ratio of power consumed in the branches $(P + Q)$ and $(R + S)$ is

In a typical Wheatstone network,the resistances in cyclic order are $A = 10 \,\Omega $,$B = 5 \,\Omega $,$C = 4 \,\Omega $ and $D = 4 \,\Omega $. What should be done for the bridge to be balanced?