Cells, EMF, Internal Resistance

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 :

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

Assertion: When a battery is short-circuited, the terminal voltage is zero.
Reason: In the situation of a short-circuit, the current is zero
 

The potential difference across the terminals of the cell when the battery is short-circuited is zero.

When a battery is short-circuited, the terminal voltage is zero

The terminal potential difference of a cell when short-circuited is (E=EMF of the 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 -

Clarify the difference in the following:

Electric conductor and voltage.

Chemical energy is converted into electrical energy with the help of _____(electric cell/electric motor).

Can the terminal voltage of cell be greater than the emf of a cell?

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 a potentiometer experiment, the balancing length of a cell is 560 cm. When an external resistance of $10\mathrm{\Omega }$ is connected in parallel to the cell, the balancing length chages by 60 cm. The internal resistance of a cell is

A daniel cell is balanced at 125 cm length of a potentiometer wire. When the cell is circuited by a resistance of $\mathrm{2 }\mathrm{\Omega }$, the balance length is obtained at 100 cm. The internal resistance of the Daniel cell is

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:

Assertion: A potential difference measured across cell with a voltmeter is always less than emf of cell.
Reason: Potentiometer is used for emf measurement.

Assertion: In a simple battery circuit, the point at the lowest potential is the positive terminal of the battery.

Reason: The current flows towards the point of the lowest potential, as it does in such a circuit from the negative to the positive terminal.

Assertion: When the cell is in the open circuit, there is no force on a test charge inside the electrolyte of the cell.

Reason: There is no field inside the cell, when the cell is in open circuit.

Assertion: In a simple battery circuit the point at the lowest potential is positive terminal of the battery.

Reason: The current flows towards the point of the lowest potential, as it does in such a circuit from negative to the positive terminal.

For a cell, the terminal potential difference is $2.2 \mathrm{V}$ when the circuit is open and reduced to $1.8 \mathrm{V}$ when the cell is connected to the resistance of $R=5 \mathrm{\Omega }$ . Determine the internal resistance of the cell.

The internal resistance of a cell of emf $2 \mathrm{V}$ is $0.1 \mathrm{\Omega }$ . It is connected to the resistance of $3.9 \mathrm{\Omega }$ . The voltage across the cell will be