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

The variation of terminal potential difference $\left(V\right)$ with current flowing through as cell is as shown

The $\mathrm{emf}$ and internal resistance of the cell are

Explain the working of Daniel cell with a neat diagram.

A battery of e.m.f $12 \mathrm{V}$ and internal resistance $0.5 \mathrm{\Omega }$ is charged by a battery charger which supplies a $132 \mathrm{V}$ d.c. supply using a series resistance of $11.5 \mathrm{\Omega }$. What is the terminal voltage of the battery during charging?

In case two cells are identical, each of emf $E=5 \mathrm{V}$ and internal resistance $r=2 \Omega$, calculate the voltage across the external resistance $R=10 \Omega$

Two cells of emf $2E$ and $E$ with internal resistance $r_1$and $r_2$ respectively are connected in series to an external resistor $R$ (see figure). The value of $R,$ at which the potential difference across the terminals of the first cell becomes zero is

Consider an electrical conductor connected across a potential difference $V.$ Let $\Delta q$ be a small charge moving through it in time $\Delta t$. If $I$ is the electric current through it,
(i) the kinetic energy of the charge increases by $IV\Delta t$
(ii) the electric potential energy of the charge decreases by $IV\Delta \mathrm{t}$.
(iii) the thermal energy of the conductor increases by $IV\Delta \mathrm{t}$.

Then the correct statements is/are

Two cells of the same EMF $E$ but different internal resistances $r_1$ and $r_2$ are connected in series with an external resistance $R$ as shown in the figure. The terminal potential difference across, the second cell is found to be zero. The external resistance $R$ must then be: