Three resistances $P, Q, R$ each of $2 \,\Omega$ and an unknown resistance $S$ form the four arms of a Wheatstone bridge circuit. When a resistance of $6 \,\Omega$ is connected in parallel to $S$,the bridge gets balanced. What is the value of $S$ in $\Omega$?

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 ............... $A$.

Assertion : Kirchhoff's junction rule follows from the conservation of charge.
Reason : Kirchhoff's loop rule follows from the conservation of momentum.

What is the current in $A$ flowing through $BD$ in the given circuit?

The total length of the potentiometer wire $AB$ is $50 \text{ cm}$ in the arrangement as shown in the figure. If $P$ is the point where the galvanometer shows zero reading, then the length $AP$ is . . . . . . $\text{ cm}$.

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: