In the circuit shown in $\text{figure} E, F, G \text{and} H$ are cells of $\text{emf} 2 \mathrm{V}, 1 \mathrm{V}, 3 \mathrm{V} \text{ and }1 \mathrm{V}$  and their internal resistances are $2 \mathrm{\Omega }, 1 \mathrm{\Omega }, 3 \mathrm{\Omega } \text{ and }1 \mathrm{\Omega }$, respectively. Calculate.

 

$\left(\mathrm{i}\right)$ The potential difference between $B \text{and} D$ is given by $\left(\frac{13-n}{13}\right)$ Volt then value of $n$ will be.

$\left(\mathrm{ii}\right)$ The ratio of potential difference across the terminals of the cell $G$ to cell $H$ is given by $\left(\frac{n+2}{19}\right)$ the value of $n$ will be.

Figure shows an arrangement to measure the emf $\mathrm{\epsilon }$and internal resistance $r$ of a battery. The voltmeter has a very high resistance and the ammeter has a very small resistance. The voltmeter reads $1.52 \mathrm{V}$ when the switch $S$ is open. When the switch is closed the voltmeter reading drops to $1.45 \mathrm{V}$ and the ammeter reads $1.0 \mathrm{A}.$ The internal resistance of the battery in $\mathrm{m\Omega }$ will be?

In the circuit shown, reading of the voltmeter connected across $400 \mathrm{\Omega }$ resistance is $60 \mathrm{V}.$ If it is connected across $300 \mathrm{\Omega }$ resistance then reading in volt will be

In the given circuit the ammeter $A_1 \text{and} A_2$ are ideal and the ammeter $A_3$ has a resistance of $1.9 \times {10}^{–3} \mathrm{\Omega }$. If sum of readings of all three meters is given by $\left(\frac{2n}{27}\right)$ Ampere the value of $n$ will be.