Potential Difference and EMF of a Cell

The reading on a high resistance voltmeter, when a cell is connected across it, is $2.0V$. When the terminals of the cell are also connected to a resistance of  $3 \Omega$  as shown in the circuit, the voltmeter reading drops to $1.5V$. The internal resistance of the cell is :

Two sources of equal emfs are connected in series. This combination is connected to an external resistance $R$. The internal resistances of the two sources are $r_1$ and $r_2\left(r_1>r_2\right)$. If the potential difference across the source of internal resistance $r_1$ is zero then the value of $R$ will be

Two identical cells each of emf $1.5 \mathrm{V}$ are connected in parallel across a parallel combination of two resistors each of resistance $20 \mathrm{\Omega }$. A voltmeter connected in the circuit measures $1.2 \mathrm{V}$. The internal resistance of each cell is :

Is emf of cell is dependent on internal resistance of cell.

A cell of $\mathrm{e}\mathrm{m}\mathrm{f}\mathrm{ }\mathrm{E}$ and internal resistance $\mathrm{r}$ is connected in series with an external resistance $\mathrm{n}\mathrm{r}$ . Then, the ratio of the terminal potential difference to $\mathrm{e}\mathrm{m}\mathrm{f}$ is -

In the shown circuit, what is the potential difference across $A$ and $B$?

The potential difference $\left(V\right)$ across the terminals of a battery which is getting charged is always:

The terminal potential difference $\left(V\right)$ of a battery of emf $\left(E\right)$ and internal resistance $\left(r\right)$ which is getting charged with electric current $\left(I\right)$ flowing through it is:

A current $2 \mathrm{A}$ flows through a $2\mathrm{\Omega }$ resistor when connected across a battery. The same battery supplied a current $0.5 \mathrm{A}$ when connected across a $9\mathrm{\Omega }$ resistor. The internal resistance of the battery is

A cell of emf $\epsilon$ and internal resistance $\left(r\right)$ is connected in series with an external resistance $\left(nr\right)$ then the ratio of the terminal potential difference to emf is:

The internal resistance of a cell is the resistance of

For a cell, the terminal potential difference is $2.2 \mathrm{V}$ when circuit is open and reduces to $1.8 \mathrm{V}$ when cell is connected to a resistance $R=5 \mathrm{\Omega }$. The internal resistance $\left(r\right)$ of the cell is

A storage battery of emf $8.0 \mathrm{V}$ with internal resistance $0.5\mathrm{\Omega }$ is being charged by a $120 \mathrm{V} \mathrm{D}\mathrm{C}$ supply using a series resistor of $15.5\mathrm{\Omega }.$ What is the terminal voltage of the battery during charging?

In the circuit shown, the potential difference between $A$ and $B$ is:

In the figure shown, if the diode forward voltage drop is 0.2 V, the voltage difference between $A$ and $B$ is

A cell has a terminal voltage $2 \mathrm{V}$ in an open circuit and the internal resistance of the given cell is $2\mathrm{ \Omega }$. If $4 \mathrm{A}$ of current is flowing between points $P$ and $Q$ in the circuit, then the potential difference between $P$ and $Q$ is

First order derivation of thermos emf produced in thermos - couple with respect to temperature gives

Two thin rings each of radius $R$ are placed at a distance  $d$ apart. The charges on the rings are $+q$ and $-q.$ The potential difference between their centres will be -

Two cells, having the same emf which are connected in series through an external resistance $R$. Cells have internal resistances ${\mathrm{r}}_1$ and ${\mathrm{r}}_2\left({\mathrm{r}}_1>{\mathrm{r}}_2\right)$ respectively. When the circuit is closed, then the potential difference across the first cell is zero. The value of $R$ is:

In the shown circuit, all three capacitor are identical and have capacitance $C \mu F$ each. Each resistor has resistance of $R \mathrm{\Omega }.$ An ideal cell of emf V volts is connected as shown. Then the magnitude of potential difference across capacitor $C_3$ in steady state is: