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. 

A battery of $\mathrm{emf} 1.4 \mathrm{V}$ and internal resistance $2 \mathrm{\Omega }$ is connected to a resistor of $100 \mathrm{\Omega }$ through an ammeter. The resistance of the ammeter is $\frac{4}{3} \mathrm{\Omega }$ A voltmeter has also been connected to find the potential difference across the resistor.

$\left(\mathrm{i}\right)$ Draw the circuit diagram.

$\left(\mathrm{ii}\right)$ The ammeter reads $0.02 \mathrm{A}.$ What is the resistance of the voltmeter ?

$\left(\mathrm{iii}\right)$ The voltmeter reads $1.10 \mathrm{V},$ what is the zero error in the voltmeter ?

(Hint : zero error $=$observed reading – actual reading)

In the figure the potentiometer wire $AB$ of length $L$ & resistance $9r$ is joined to the cell $D$ of $\mathrm{e}.\mathrm{m}.\mathrm{f}. \mathrm{\epsilon }$ & internal resistance $r$. The cell $C\text{'}$s e.m.f. is $\frac{\mathrm{\epsilon }}{2}$and its internal resistance is $2r.$ The galvanometer $G$ will show no deflection then find length $AJ$

Figure shows a $2.0 \mathrm{V}$ potentiometer used for the determination of internal resistance of $1.5 \mathrm{V}\text{ cell.}$ The balance point of the cell without $9.5 \mathrm{\Omega }$in the external circuit is $70 \mathrm{cm}.$ When a resistor of $9.5 \mathrm{\Omega }$ is used in the external circuit of the cell, the balance point shifts to $60 \mathrm{cm}$ length of the potentiometer wire. Determine the internal resistance of the secondary cell.

Figure shows a potentiometer with a cell of emf $2.0 \mathrm{V}$ and internal resistance $0.04 \mathrm{\Omega }$ maintaining a potential drop across the potentiometer wire$\mathrm{AB}.$A standard cell which maintains a constant emf of $1.02 \mathrm{V}$ (for very moderate currents up to a few ampere) gives a balance point of $67.3 \mathrm{cm}$ length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of $600 \mathrm{k\Omega }$ is put in series with it which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf E and the balance point found similarly turns out to be at $82.3 \mathrm{cm}$ length of the wire.

$\left(\mathrm{a}\right)$ What is the value of $E$ ?                                                                                                                            

$\left(\mathrm{b}\right)$What purpose does the high resistance of $600 \mathrm{k\Omega }$ have ?

$\left(\mathrm{c}\right)$ Is the balance point affected by this high resistance?

$\left(\mathrm{d}\right)$ Is the balance point affected by the internal resistance of the driver cell?

$\left(\mathrm{e}\right)$ Would the method work in the above situation if the driver cell of the potentiometer had an emf of $1.0 \mathrm{V}$ instead of $2.0 \mathrm{V}?$

$\left(\mathrm{f}\right)$ Would the circuit work well for determining extremely small emf, say, of the order of few mV (such typical emf of thermocouple)?

Figure shows a metre bridge (which is nothing but a practical Wheatstone Bridge) consisting of two resistors $\mathrm{X} \text{and} \mathrm{Y}$ together in parallel with a metre long constantan wire of uniform cross-section. With the help of a movable contact $\mathrm{D}$, one can change the ratio of the resistances of the two segments of the wire until a sensitive galvanometer G connected across $\mathrm{B} \text{and} \mathrm{D}$shows no deflection. The null point is found to be at a distance of $30 \mathrm{cm}$ from the end $\mathrm{A}$. The resistor$\mathrm{Y}$ is shunted by a resistance of $12.0 \mathrm{\Omega }$ and the null point is found to shift by a distance of $10 \mathrm{cm}.$ Determine the resistance of $\mathrm{X} \text{and} \mathrm{Y}\text{.}$

Connect a battery to the terminals and complete the circuit diagram so that it works as a potential divider meter. Indicate the output terminals also.

The drift velocity of electrons in a conducting wire is of the order of $1 \mathrm{mm} {\mathrm{s}}^{-1}$, yet the bulb glows very quickly after the switch is put on because