Solution:

Ans. (i) Try yourself Similar to Q. 1, Short Answer Type Questions-I

Try yourself Similar to Q. 2, Short Answer Type Questions-I

(ii)

$$\begin{array}{rlrlrlrlr}& \frac{R_1}{R_2}=\frac{40}{60}=\frac{1}{3}& \frac{R_1+10}{R_2}=\frac{60}{40}=\frac{3}{2}& \frac{R_1}{R_2}+\frac{10}{R_2}=\frac{3}{2}& \Rightarrow \frac{2}{3}+\frac{10}{R_2}=\frac{3}{2}& R_2=12\mathrm{\Omega }& \Rightarrow \text{ Substituting for }{R}_2\text{ and finding the value of }{R}_1& R_1=8\mathrm{\Omega }& \text{ [CBSE Marking Scheme, 2013] }\end{array}$$

[AI Q. 4. (i) Obtain the condition under which the current flowing, in the current detecting device, used in the circuit shown in the figure, becomes zero.

(ii) Describe briefly the device, based on the above question. Draw a circuit diagram for this device and discuss, in brief, how it is used for finding an unknown resistance. U [Foreign Comptt. 2016]

Ans. (i) The given circuit can be redrawn as shown :

It is, therefore, a Wheatstone Bridge. $1/2$ Using Kirchhoff’s laws, we get (when $i_g=0$ )

$\begin{array}{rlr}{i}_l& =i_3\text{ and }{i}_2& =i_4\end{array}$

For the loop $ABDA$, we have

$-i_{1}P+i_{2}X=0$ or $i_{1}P=i_{2}X$

For the loop $BCDB$ we have,

$-i_{3}Q+i_{4}R=0$ or $i_{3}Q=i_{4}R$

Dividing we get,

$$\begin{array}{rlr}\frac{i_{1}P}{i_{3}Q}& =\frac{i_{2}X}{i_{4}R}\text{ or }\frac{P}{Q}& =\frac{X}{R}{(\because i_1=i_3\text{ and }{i}_2=i_4)}^{1/2}\end{array}$$

(ii) A simple device, based on the above condition, is the metre bridge. It has a (uniform cross-section) wire of length $1\mathrm{m}$ stretched out between two thick metallic clamps. It has two gaps for connection a resistance box and the unknown resistance. $1/2$ The circuit diagram, for the metre bridge, is shown here.

We move the jockey, on the wire of the meter bridge, till we find a point at which the deflection in $\mathrm{G}$, is zero. we can have

$$\frac{R}{l_1}=\frac{X}{l_2}$$

or

$$S=R\left(\frac{l_2}{l_1}\right)$$

Knowing $R$, and finding $l_1$ and $l_2=(100-l_1)$, we can easily calculate $X$.

[CBSE Marking Scheme 2016]

TOPIC-3

Metre Bridge, Potentiometer and their Applications

Revision Notes

Metre Bridge

D It is an instrument which is used to find the unknown resistance of a coil or a material connected in a circuit.

• It is also known as slide wire bridge which is an instrument that works on the principle of Wheatstone bridge.

Metre bridge has two metallic strips which act as holders for the wire that are made of metals like copper.

$>$ In metre bridge :

• Resistance box $R_B$ and unknown resistance $R$ are connected across the two gaps of metallic strips.
• One end of galvanometer is connected to the middle lead of metallic strip placed between $L$ shaped strips while other end is connected to a jockey.
• Jockey which is a metal wire having one end as knife edge is used for sliding on the bridge wire.

Measurement from the Metre bridge :

At negative terminal of galvanometer, there appears zero deflection that makes jockey to connect to negative point on the wire.

• The distance from point $X$ to $Y$ is taken as $\mathrm{L}{1} \mathrm{~cm}$whilethedistancefrompoint$Y$topoint$Z$istakenas$L{2} \mathrm{~cm}$whichcanbe$\left(100-\mathrm{L}_{1}\right) \mathrm{cm}$.

Metre bridge can be drawn similar to Wheatstone bridge as :

Metre bridge

Wheastone bridge

From the above arrangement :

$$\begin{array}{rlrlrl}& \frac{R_B}{\text{ Resistance across }XY}=\frac{R}{\text{ Resistance across }YZ}& \text{ Now, }\frac{R_B}{\frac{\rho L_1}{A}}=\frac{R}{\frac{\rho L_2}{A}}& \text{ Further, }& \frac{R_B}{\frac{\rho L_1}{A}}=\frac{R}{\frac{\rho (100-L_1)}{A}}& \frac{R_B}{L_1}=\frac{R}{100-L_1}\end{array}$$

$$\begin{array}{r}[\text{ As, }R=\frac{\rho L}{A}][\because L_2=100-L_1]\end{array}$$

Potentiometer

P Potentiometer is a device which measures the emf of particular cell and helps in comparing the emfs of different cells.

P Potentiometer depends on deflection method where zero deflection results in non drawn of current from the cell or circuit.

It serves as an ideal instrument of infinite resistance for measuring the potential difference.

Potentiometer comprises of long resistive wire $AB$ of length $L$ (about $6\mathrm{m}$ to $10\mathrm{m}$ long) made up of manganin or constantan.

In this, a battery of known voltage $E$ and internal resistance $r$ forms the primary circuit.

In the potentiometer circuit, one terminal of other cell is connected at one end of main circuit while other terminal is connected at any point on the resistive wire through galvanometer $G$ which forms the secondary circuit.

where, $J=$ Jockey, $K=$ Key, $R_h=$ Variable resistance which controls the current through the wire $AB$

In the circuit :

Specific resistance ( $\rho$ ) of wire is high while its temperature coefficient of resistance is low.

• At point $A$, all high potential points of primary and secondary circuits are connected together, while all low potential points are connected to point $B$ or jockey.
• Value of known potential difference is more than the value of unknown potenthal difference that is to be measured.

The current in primary circuit should remain constant and jockey shouldnot shide with the wire.

Principle of Potentiometer

Potentiometers are displacement sensors that produce electrical butput in proportion to the mechanical displacement.

• It can be used to measure the internal resistance and emf of a cell which cannot be measured by voltmeter.

The basic principle of potentiometer is that the potential drop along any length of the wire is directly proportional to its length. So when a constant current flows through w wire of uniform cross-section and composition then,

When there is zero potential difference between two points, there will be no flow of electric current.

• Applications of Potentiometer : in measuring potential difference and comparing emf of cells Potentiometer in measuring potential difference.
• In a potentiometer auxiliary circuit comprised of battery of emf $E$ connected across terminals $A$ and $B$ with rheostat $R_h$, resistance box and key $K^{\text{in }}$ series where resistance $R_1$ is connected to terminal $A$ and jockey $J$ through galvanometer with cell $E_1$ and key $K_1$ in series, then if key $K_1$ is closed, current will flow through resistance $R_1$ where a potential difference is developed.
• If $\mathrm{J}$ is the position of jockey onpotentiometer wire which gets adjusted in such a way that galvanometer shows no deflection, then $AJ$ will be the balancing length $l$ on potentiometer wire.
• Here, the Galvanometer will show no deflection as potential is same if key $K$ is potential gradient of potentiometer wire, then potential difference across resistance $R_1$ will result as :

$$V=Kl$$

If $r$ is the resistance of potentiometer of length $L$, then current through potentiometer will be :

$$I=\frac{E}{R+r}$$

Potential drop across potentiometer wire will be :

$$\mathrm{Ir}\text{ or }\left(\frac{E}{R+r}\right)\times r$$

Now potential gradient of potentiometer wire which is potential per unit length is :

$$\begin{array}{rlr}& K=\left(\frac{E}{R+r}\right)\times \frac{r}{L}\therefore & V=\left(\frac{E}{R+r}\right)\times \frac{rl}{L}\end{array}$$

Application of Potentiometer in comparing emf of cells

If a positive terminal of the cell of emf $E_1$ is connected to terminal $A$ while negative terminal is connected to jockey by galvanometer, then by closing the key, jockey will move along the wire $AB$ and null point is obtained where galvanometer shows no deflection.

When the length of wire $AP$ as $l_1$ is measured, then potential difference across it, will balance the emf $E_1$, So $E_1=K{l}_1$, where $K$ is potential gradient of the wire.

When cell of emf $E_1$ is disconnected while cell of emf $E_2$ is connected, then $E_2=K{l}_2$.

On comparing and dividing we get an expression :

$$\frac{E_1}{E_2}=\frac{l_1}{l_2}$$

By knowing the values of $l_1$ and $l_2$, the emf of two cells can be compared.

Applications of potentiometer in measurement of internal resistance of gell

(i) Initially in secondary circuit key $K$ ’ remains open and the balancing length $\left(l_1\right)$ is obtained. Since cell $\mathrm{E}$ is in open circuit so its emf balances on length $l_1$, i.e. $E=K{l}_1$

${\mathrm{R}}_h$

(ii) Now key $K^{\prime }$ is closed so cell $E$ comes in closed circuit. If the process is repeated again then the potential difference $V$ balances on length $l_2$, i.e

$$\begin{array}{}\text{(b)}& V=K{l}_2\end{array}$$

(iii) By using formula, internal resistance, $r=(\frac{E}{V}-1)\cdot R^{\prime }$

$$r=\left(\frac{l_1-l_2}{l_2}\right)\cdot R^{\prime }$$

Know the Formulae

Potential gradient $(K)$ :

$$\begin{array}{rlr}& K=\frac{V}{L}=\frac{iR}{L}=\left(\frac{E}{R+R_h+r}\right)\times \frac{R}{L}& r=(\frac{E}{V}-1)\times R=\left(\frac{l_1-l_2}{l_2}\right)\times R\end{array}$$

Internal resistance of a cell :

Comparison of emf’s of two cells : $\frac{E_1}{E_2}=\frac{l_1}{l_2}$

Objective Type Questions