The current drawn from the battery in the given network is (Internal resistance of battery is neglected) (in $A$)

Why is the circuit configuration known as a Wheatstone bridge?

In a Wheatstone bridge,$P = 90\,\Omega$,$Q = 110\,\Omega$,$R = 40\,\Omega$,and $S = 60\,\Omega$. $A$ cell of $4\,V$ emf is connected across the input terminals. The potential difference between the diagonal points $B$ and $C$ (where the galvanometer is connected) is ............. $V$.

The equivalent resistance between $A$ and $B$ in the given circuit is . . . . . . . . (in $\Omega$)

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

$A$ Wheatstone's bridge is balanced with a resistance of $625\, \Omega$ in the third arm,where $P, Q$ and $S$ are in the $1^{st}, 2^{nd}$ and $4^{th}$ arm respectively. If $P$ and $Q$ are interchanged,the resistance in the third arm has to be increased by $51\,\Omega$ to secure balance. The unknown resistance in the fourth arm is ............. $\Omega$.