The sliding contact $C$ is at one-fourth of the length of the potentiometer wire $(AB)$ from $A$ as shown in the circuit diagram. If the resistance of the wire $AB$ is $R_0$,then the potential drop $(V)$ across the resistor $R$ is

An ideal cell of emf $10\, V$ is connected in the circuit shown in the figure. Each resistance is $2\, \Omega$. The potential difference (in $V$) across the capacitor when it is fully charged is

$A$ uniform metallic wire having resistance $4 \ \Omega$ is bent to form a square loop $(ABCD)$ (see figure). $A$ resistance of $2 \ \Omega$ is connected between points $B$ and $D$ and a battery of $2 \ V$ is connected across points $A$ and $C$ as shown in the figure. Now the value of current $(I)$ is: (in $A$)

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:

$A$ 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 circuit shown here, what is the value of the unknown resistor $R$ so that the total resistance of the circuit between points $P$ and $Q$ is also equal to $R$?