Consider the circuit shown in figure. Find the current through the $10 \mathrm{\Omega }$ resistor when the switch $\mathrm{S}$ is

$\left(\mathrm{a}\right)$opened

$\left(\mathrm{b}\right)$closed.

Six lead-acid type of secondary cells, each of emf $2.0 \mathrm{V}$ and internal resistance $0.015 \mathrm{\Omega }$, are joined in series to provide a supply to a resistance of $8.5 \mathrm{\Omega }$. Determine : (i) the current drawn from the supply and (ii) its terminal voltage.

In the figure each cell has an emf of $1.5 \mathrm{V}$ and internal resistance of $0.40 \mathrm{\Omega }$. Calculate:

(i) current $\mathrm{I}$

(ii) current in the $36 \mathrm{\Omega }$ resistor

(iii) potential difference across $\mathrm{A}$ and $\mathrm{B}$.

In the circuit shown all five resistors have the same value $200 \mathrm{\Omega }$ and each cell has an emf $3 \mathrm{V}$. Find the open circuit voltage and the short circuit current for the terminals $\mathrm{A}$ and $\mathrm{B}$.

Find the currents through the three resistors shown in figure

Find the value of $\frac{i_1}{i_2}$ in figure if (a) $R= 0.1 \mathrm{\Omega }$, (b) $R=1 \mathrm{\Omega }$ (c) $R= 10 \mathrm{\Omega }$. Note from your answer that in order to get more current from a combination of two batteries they should be joined in parallel if the external resistance is small and in series if the external resistance is large as compared to the internal resistances.

A voltmeter of resistance $400 \mathrm{\Omega }$ is used to measure the potential difference across the $100 \mathrm{\Omega }$ resistor in the circuit shown in the figure.

(a) What will be the reading of the voltmeter?

(b) What was the potential difference across $100 \mathrm{\Omega }$ before the voltmeter was connected? 

An electrical circuit is shown in the figure. Calculate the potential difference across the resistance of $400 \mathrm{ohm},$ as will be measured by the voltmeter $V$ of resistance $400 \mathrm{ohm},$either by applying Kirchhoff's rules or otherwise.