What is the internal resistance of the cell if the e.m.f. of the cell is $2.08\hspace{0.17em}\mathrm{volt}$ and the terminal potential of the cell is $2.00 \mathrm{volt}$ when a $2 \Omega$ resistance is connected?
 

The figure shows a part of a closed circuit. If the current flowing through it is $2 \mathrm{A}$ then $V_B-V_A$ is _______

When an ammeter of negligible internal resistance is inserted in series with the circuit, it reads $1 \mathrm{A}$. When a voltmeter of very large resistance is connected across $R_1$, it reads $3 \mathrm{V}$. But when the points $A$ and $B$ are short-circuited by a conducting wire, then the voltmeter measures $10.5 \mathrm{V}$ across the battery. The internal resistance of the battery is equal to

The internal resistances of two cells shown are $0.1 \mathrm{\Omega }$ and $0.3 \mathrm{\Omega }$. If $R=0.2 \mathrm{\Omega }$, the potential difference across the cell

A voltmeter having a resistance of $1800 \mathrm{\Omega }$ is employed to measure the potential difference across $200 \mathrm{\Omega }$ resistance, which is connected to dc power supply of $50 \mathrm{V}$ and internal resistance $20 \mathrm{\Omega }$. What is the approximate percentage change in the potential difference across $200 \mathrm{\Omega }$ resistance as a result of connecting the voltmeter across it?

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 battery has an $\mathrm{e}\mathrm{m}\mathrm{f}$ of $15\mathrm{ }\mathrm{V}$ and internal resistance of $1 \mathrm{\Omega }$ . The terminal to terminal potential difference is less than, equal to or greater than $15\mathrm{ }\mathrm{V}$, if the current in the battery is

(1) from negative to positive terminal,

(2) from positive to negative terminal,

(3) zero current?