Battery

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 :

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 $\alpha$.

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

The terminal voltage of a cell in open circuit condition _____is its emf.

A potential difference across the terminals of a battery is $50 \mathrm{V}$ when $11 \mathrm{A}$ current is drawn and $60 \mathrm{V}$ when $1 \mathrm{A}$ current is drawn. The emf and the internal resistance of the battery are

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 -

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:

Two nonideal batteries are connected in parallel. Consider the following statemnents:
(a) The equivalent emf is smaller than either of the two emfs
(b) The equivalent internal resistance is smaller than either of the two internal resistance

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

In the given circuit reading of ideal voltmeter is $\frac{E}{2}$ ,then internal resistance of cell is :

A battery has an emf $15$ volt and internal resistance $1 \mathrm{\Omega }$ Is the terminal to terminal potential difference less than, equal to or greater than $15$ volt if the current in the battery is:

(i) from $\left(-\right)ve$ to $\left(+\right)ve$ terminal

(ii) from $\left(+\right)ve$ to $\left(-\right)ve$ terminal

(iii) zero current

The efficiency of a cell when connected to a resistance $R$ is $60\%.$ What will be its efficiency if the external resistance is increased to six times.

A battery of emf $10 \mathrm{V}$ and internal resistance $3 \mathrm{\Omega }$ is connected to a resistor. If the current in the circuit is $0.5 \mathrm{A}$, what is the resistance of the resistor

If a cell is flowing current through external resistance then curve between terminal potential $\left(V\right)$ and external resistance $R$ 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: