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Derive an expression for resistance measurement using Wheatstone Bridge.

Mumbai University > Electronics and Telecommunication > Sem3 > Electronic Instruments and Measurements

Marks: 6M

Year: Dec 13

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For measuring any electrical resistance accurately, Wheatstone bridge is widely used. There are two known resistors - one variable resistor and one unknown resistor connected in bridge form as shown below. By adjusting the variable resistor, the current through the Galvanometer is made zero. When the current through the galvanometer becomes zero, the ratio of two known resistors is exactly equal to the ratio of adjusted value of variable resistance and the value of unknown resistance. In this way, the value of unknown electrical resistance can easily be measured by using a Wheatstone Bridge.

Working

The general arrangement of Wheatstone bridge circuit is shown in the figure below. It is a four arms bridge circuit where arm AB, BC, CD and AD are consisting of electrical resistances P, Q, S and R respectively. Among these resistances P and Q are known fixed electrical resistances and these two arms are referred as ratio arms.

An accurate and sensitive Galvanometer is connected between the terminals B and D through a switch $S_2$. The voltage source of this Wheatstone bridge is connected to the terminals A and C via a switch $S_1$ as shown. A variable resistor S is connected between point C and D

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The potential at point D can be varied by adjusting the value of variable resistor. Suppose current $I_1$ and current $I_{(2)}$are flowing through the paths ABC and ADC respectively.

If we vary the electrical resistance value of arm CD the value of current $ I_2$ will also be varied as the voltage across A and C is fixed. If we continue to adjust the variable resistance one situation may come when voltage drop across the resistor S that is $I_2$.S, which is exactly equal to voltage drop across resistor Q that is $I_1$.Q

Thus the potential at point B becomes equal to the potential at point D hence potential difference between these two points is zero; hence, current through galvanometer is nil. Then the deflection in the galvanometer is nil when the switch $S_2$ is closed.

Now, from Wheatstone bridge circuit, we get

And

    Current $I_2=\frac{V}{R+S}$

Now potential of point B in respect of point C is nothing but the voltage drops across the resistor Q and this is

    $I_1Q=\frac{V.Q}{P+Q}$---------------------(i)

Again potential of point D in respect of point C is nothing but the voltage drops across the resistor S and this is

    $I_1Q=\frac{V.S}{R+S}$------------------------(ii)

Equating equations (i) and (ii) we get,

    $\frac{V.Q}{V+Q}=\frac{V.S}{R+S}-\gt\frac{Q}{P+Q}=\frac{S}{R+S}$

    $-\gt\frac{P+Q}{Q}=\frac{R+S}{S}-\gt\frac{P}{Q}+1=\frac{R}{S}+1-\gt\frac{P}{Q}=\frac{R}{S}$

    $-\gt R=S\times \frac{P}{Q}$

Here in the above equation, the value of S and $\frac{P}Q $ are known, so value of R can easily be determined.

The electrical resistances P and Q of the Wheatstone bridge are made of definite ratio such as 1:1; 10:1 or 100:1 known as ratio arms and S the rheostat arm is made continuously variable from 1 to 1,000 Ω or from 1 to 10,000 Ω.

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