Two identical bulbs are connected in parallel across an ideal source of emf $E$. The ammeter $A$ and voltmeter $V$ are ideal. If bulb $B_2$ gets fused,then

An electrical circuit consists of ten $100 \,\Omega$ resistors. Out of these $10$ resistors,a group of $n_1$ resistors are connected in parallel and another group of $n_2$ resistors are separately connected in parallel. These two groups are then connected in series and this combination is connected to a voltage source of $100 \,V$. If the net current through the circuit is $2.5 \,A$,the values of $n_1$ and $n_2$ are:

When three identical bulbs of $60 \;W, 200 \;V$ rating are connected in series to a $200 \;V$ supply,the power drawn by them will be ....... $Watt$.

Under what condition can the current passing through the resistance $R$ be increased by short-circuiting the battery of emf $E_2$? The internal resistances of the two batteries are $r_1$ and $r_2$ respectively.

Two wires of thickness $1 \, mm$ and $2 \, mm$ have the same mass of copper. The two wires are connected in series and a current is passed through them. The ratio of the power generated in the wires will be ....

$A$ copper voltameter is connected to a $12 \, V$ source,and $2 \, g$ of copper is deposited in $30 \, min$. If this cell is connected to a $6 \, V$ source,how much copper (in $g$) will be deposited in $45 \, min$?