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• Sunil Rawat

# Which aircraft system use the wheatstone bridge principle ?

The Wheatstone bridge is used for temperature measurement, where the temperature probe is connected in place of the unknown resistance, and the circuit is balance using the variable resistance, so the indicator reads zero when the temperature of the probe is zero.

Whenever the temperature changes, the resistance of the probe changes which cause an unbalance condition of the Wheatstone bridge, and the current flows through the galvanometer. The galvanometer is replaced by a temperature indicator which is basically an ammeter calibrated in terms of temperature.

Let us look at the basic wheatstone bridge.

## Definition:

The device uses for the measurement of minimum resistance with the help of comparison method is known as the Wheatstone bridge. The value of unknown resistance is determined by comparing it with the known resistance.

The Wheatstone bridge works on the principle of null deflection, i.e. the ratio of their resistances are equal, and no current flows through the galvanometer. The bridge is very reliable and gives an accurate result.

In unbalanced condition, the current flow through the galvanometer. When zero current passes through the galvanometer, then the bridge is said to be in balanced condition.

This can be done by adjusting the known resistance P, Q, R and the variable resistance S.

## Construction of Wheatstone Bridge

The basic circuit of the Wheatstone bridge has four arms which consist two unknown resistance, one variable resistance and the one unknown resistance along with the emf source and galvanometer.

The emf supply is attached between point A and C, and the galvanometer is connected between point B and D. The current through the galvanometer depends on the potential difference across it.

## Working of Galvanometer

Let the total current flow in the circuit be I, at junction A the current splits into I1 and I2. at junction A, the current I1 splits into I3 and IG, at junction D the current IG and I2 combine to form current I4 and at junction C the current I3 and current I4 combine to the total current I. P, Q, R, S be the resistance and G be the resistance of the galvanometer.

Now if we apply Kirchoff's current law, which state that current entering the junction is equal to current leaving the junction.

At junction B,

I1 = I3 + IG

At junction D,

IG + I2 = I4

Now if we apply Kirchoff's voltage laws, which states that the voltage drop across a closed loop is always equal to zero.

For loop ABD,

I1*P + IG*G - I2*R = 0 (the direction of I2 is opposite to I1 and IG hence the negative sign)

I1*P + IG*G = I2*R

For loop BCD, I3*Q - I4*S - IG*G = 0 (the direction of I4 and IG is opposite to I3 hence the negative sign) I3*Q = I4*S + IG*G

Since we know at balance condition the current flow through the galvanometer is zero. Hence current IG is zero.

At junction B, I1 = I3 + IG I1 = I3 (Since IG = 0) ------------- 1

At junction D, IG + I2 = I4 I2 = I4 (Since IG = 0) ------------- 2

Also, For loop ABD, I1*P + IG*G = I2*R I1*P = I2*R (Since IG = 0) ------------- 3

For loop BCD, I3*Q= I4*S + IG*G I3*Q = I4*S (Since IG = 0) ------------- 4 By dividing equation 3 and 4 I1*P/I3*Q = I2*R/I4*S

Substituting the values of from equation 1 and 2, we get P/Q=R/S S = R(Q/P) Other method to solve the Wheatstone bridge is using voltage divider method and since no current flows through the galvanometer, same current flow from junction A to junction C via junction B, similarly same current flow from junction A to junction C via junction D.

Understanding using Kirchoff's is much simpler method.