$A$ cell of $emf$ $E$ and internal resistance $r$ is connected in series with an external resistance $nr$. Then,the ratio of the terminal potential difference to $emf$ is

$B_{1}$, $B_{2}$ and $B_{3}$ are three identical bulbs connected to a battery of steady $EMF$ with key $K$ closed. What happens to the brightness of the bulbs $B_{1}$ and $B_{2}$ when the key $K$ is opened?

$A$ cell of emf $1.8 \ V$ gives a current of $17 \ A$ when directly connected to an ammeter of resistance $0.06 \ \Omega$. The internal resistance of the cell is: (in $\Omega$)

In the given circuit,the internal resistance of the $18\,V$ cell is negligible. If $R_1 = 400\,\Omega$,$R_3 = 100\,\Omega$,and $R_4 = 500\,\Omega$,and the reading of an ideal voltmeter across $R_4$ is $5\,V$,then the value of $R_2$ will be ........... $\Omega$.

$A$ constant voltage of $50 \ V$ is maintained between the points $A$ and $B$ of the circuit shown in the figure. The current through the branch $CD$ of the circuit is (in $A$)

Find the current in the loop.