The current in the arm $CD$ of the circuit will be

When the value of $R$ in the balanced Wheatstone bridge, shown in the figure, is increased from $5 \Omega$ to $7 \Omega$, the value of $S$ has to be increased by $3 \Omega$ in order to maintain the balance. What is the initial 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 ............... $A$.

For the given circuit,the equivalent resistance between $A$ and $B$ is:

The potential difference $V$ across the filament of the bulb shown in the given Wheatstone bridge varies as $V = i(2i + 1)$,where $i$ is the current in ampere through the filament of the bulb. The emf of the battery $(V_b)$ so that the bridge becomes balanced is (in $V$)

$A$ $6 \, V$ cell with $0.5 \, \Omega$ internal resistance, a $10 \, V$ cell with $1 \, \Omega$ internal resistance, and a $12 \, \Omega$ external resistance are connected in parallel. The current (in ampere) through the $10 \, V$ cell is