Four resistances $40 \ \Omega, 60 \ \Omega, 90 \ \Omega$,and $110 \ \Omega$ form the arms of a quadrilateral $ABCD$. $A$ battery of emf $40 \ V$ with negligible internal resistance is connected across $AC$. Find the potential difference across $BD$. (in $V$)

When a resistance $R_1$ is connected across a cell,the current is $I_1$ and if the resistance $R_1$ is replaced by $R_2$,the current is $I_2$. Then the internal resistance of the cell is

If two bulbs of wattage $25\, W$ and $100\, W$ respectively,each rated at $220\, V$,are connected in series with a supply of $440\, V$,then which bulb will fuse?

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:

$A$ $100\, W$ bulb $B_1$,and two $60\, W$ bulbs $B_2$ and $B_3$,are connected to a $250\, V$ source,as shown in the figure. Now $W_1, W_2$ and $W_3$ are the output powers of the bulbs $B_1, B_2$ and $B_3$,respectively. Then

The figure shows a network of resistances in which the point $A$ is earthed. The current through the $3\,\Omega$ resistor is