Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

WHEATSTONE BRIDGE

For a certain adjustment of Q, V_BD = 0, then no current flows through the galvanometer.

$\Rightarrow$ V_B = V_D or V_AB= V_AD $\Rightarrow$ I₁.P = I₂.R

Likewise,$V_{BC}=V_{DC}$ $\Rightarrow$ I₁.Q = I₂.S

Dividing, we get, $\frac{P}{Q}=\frac{R}{S}$

GROUPING OF IDENTICAL CELLS

1. Series Grouping

emf of the cell is e and internal resistance is r. R is the external resistance, I is the current

passing through the circuit and n is the total number of cells.

Applying Kirchhoff's law

$\epsilon$ - ir + $\epsilon$ - ir + ........ (to n times ) - iR = 0

$\Rightarrow$ $i=\frac{n\epsilon }{R+nr}$

2. Parallel grouping

Applying Kirchhoff's law

$\epsilon -\frac{I}{n}r-IR=0$

$\Rightarrow$ $I=\frac{n\epsilon }{r+nR}$

3. Mixed grouping

Number of rows is m and number of cells in each row is n

Apllying kirchhoff's law

$n\epsilon -n\frac{I}{m}r-IR=0$

$\Rightarrow$ I = $\frac{mn\epsilon }{mR+nr}$

For current through R to be maximum,

mR = nr

$\Rightarrow$ R = $\frac{nr}{m}$ = one line total internal resistance/No. of Line

AMMETER

An Ammeter is an instrument used for measuring current in electrical circuits. A galvanometer is a low resistance instrument. A large current passing through it may damage the instrument

Changing the range of an ammeter

Suppose the ammeter gives full scale deflection when a current I_g flows through it. Now if we want to convert the reading of the ammeter in such a manner that it gives full scale deflection for a higher current I in the branch of the circuit, we connect a small resistance S in parallel to the coil of the galvanometer, which has a resistance G.

The resistance value is so chosen that out of the total current I only I_g flows through the coil and the remaining current flows through S. As potential difference across S = potential difference across G.

$\Rightarrow$ (I – I_g)S = I_gG $\Rightarrow$ S = $\left(\frac{I_{g}G}{I-I_g}\right)$

So, effectively this ammeter will measure current up to I ampere and its effective resistance

= $\left(\frac{GS}{S+G}\right)$.

In practice G is large as compared to S. Therefore the effective resistance of the ammeter equals R_A $\approx$ S, which is small. An ideal ammeter has zero resistance.

VOLTMETER

A voltmeter is an instrument used for measuring potential difference across the two ends of a current carrying conductor. It is connected in parallel with the conductor across which the potential difference is to be measured. The current through the conductor should not change on connecting the voltmeter, and so the voltmeter should draw a very small current, i.e. its resistance has to be high.

When a galvanometer is used to measure potential difference across the ends of a current carrying conductor, a high resistance R is connected in series with the galvanometer.

Consider the diagram shown. Suppose the galvanometer gives full scale deflection when a current I_g passes through its coil. If G is resistance of the galvanometer coil then:

Potential difference to be measured = V = $I_g(G+R)$ $\Rightarrow$ R = $\frac{V}{I_g}$- G

So, effectively the voltmeter has resistance = R_v = (G + R)

In practice R_v is very large compared to G. An ideal voltmeter should possess infinite resistance.

POTENTIOMETER

A potentiometer is used to compare electromotive forces of two cells or to measure the internal resistance of a cell.

Principle:

The potentiometer is based upon the principle that when a constant current is passed through a wire of uniform cross sectional area, the potential drop across any portion of the wire is directly proportional to its length.

Consider the network shown in the diagram :

Here, $\left(V_A-V_B\right)=\left(\frac{E}{R_1+R_2+R}\right)R_1$

Now, let us consider that an EMF source having EMF same as (V_A – V_B) calculated above and internal resistance 'r' connected in parallel to R₁.

Suppose that a current 'I₂' passes through the EMF source

We have, applying Kirchoff's rules, the following equations

R₂ (I₁+I₂) + R₁ I₁ + R (I₁+ I₂) = E . . . (i)

R₁ I₁ = I₂ r + E¢ . . . (ii)

Solving (i) & (ii) we get

I₂ = $\frac{E{R}_1-E^{\prime }(R_1+R_2+R)}{R_1\left[R+R_2\right]+r\left[R_1+R_2+R\right]}$

For the chosen value of E' that is $\left(\frac{E}{R_1+R_2+R}\right)R_1$ we get I₂ = 0

Thus, taking a length of uniform resistance wire between A and C instead of the two resistors R₁ & R₂, it is possible to get zero (null) deflection in the galvanometer in the circuit shown below, provided that.

E₁ < IR_AC

Let ${\ell }_1$ be the length AB of the wire, corresponding to zero deflection in galvanometer, and let B be the point where the movable pointer 'P' (also called Jockey) makes contact with wire AC. If the same experiment is repeated but with another cell of EMF E₂, we will be getting a different length '${\ell }_2$' at the instant galvanometer shows null deflection. If in both the cases the rheostat 'R' is kept unchanged,

$\frac{E_1}{E_2}=\frac{{\ell }_1}{{\ell }_2}$.

The arrangement shown in figure (iii) is called Potentiometer and we use it to measure EMF using the above relationship.

Application :

A potentiometer can be used to compare emfs of two cells or to measure internal resistance of a cell. The method to compare the emfs of two cells has already been explained. When we want to measure the internal resistance of a cell, consider the following formula :

r = $\left(\frac{E}{V}-1\right)S=\left(\frac{{\ell }_1-{\ell }_2}{{\ell }_2}\right)S$,

where r = internal resistance of the cell, E = EMF, V = potential difference across the cell during current flow, S = resistance of resistance box, ${\ell }_1$ , ${\ell }_2$ are two balancing lengths.

Example : A potentiometer wire of length 100 cm has a total resistance of 10$\Omega$. It is connected in series with a resistance R and a cell of emf 2 volts and of negligible internal resistance. A cell of emf 10 mV is balanced against a length of 40cm of potentiometer wire. What is the value of the external resistance R ?

Solution : As shown in the figure, if R is the unknown resistance, the current in the circuit

I = $\frac{V}{(r+R)}=\frac{2}{(10+R)}$.

Now as the 100 cm wire has a resistance of 10 $\Omega$, the resistance of 40 cm of wire will be 40 x (10/100) = 4 ohm.

Potential drop across 40 cm wire will be V = I x4

but here V = 10 mv (given)

Hence, $10\times {10}^{-3}$ = $\frac{2}{(10+R)}\times 4$

i.e. R = 790 $\Omega$.