Question

steps to make a wheatstone bridge..

Answer 1

When balanced, the Wheatstone bridge can be analysed simply as two series strings in parallel. In our tutorial about Resistors in Series, we saw that each resistor within the series chain produces an IR drop, or voltage drop across itself as a consequence of the current flowing through it as defined by Ohms Law. Consider the series circuit below.


As the two resistors are in series, the same current ( i ) flows through both of them. Therefore the current flowing through these two resistors in series is given as: V/RT.

I = V ÷ R = 12V ÷ (10Ω + 20Ω) = 0.4A

The voltage at point C, which is also the voltage drop across the lower resistor, R2 is calculated as:

VR2 = I × R2 = 0.4A × 20Ω = 8 volts

Then we can see that the source voltage VS is divided among the two series resistors in direct proportion to their resistances as VR1 = 4V and VR2 = 8V. This is the principle of voltage division, producing what is commonly called a potential divider circuit or voltage divider network

Now if we add another series resistor circuit using the same resistor values in parallel with the first we would have the following circuit.



As the second series circuit has the same resistive values of the first, the voltage at point D, which is also the voltage drop across resistor, R4 will be the same at 8 volts, with respect to zero (battery negative), as the voltage is common and the two resistive networks are the same.

But something else equally as important is that the voltage difference between point C and point D will be zero volts as both points are at the same value of 8 volts as: C = D = 8 volts, then the voltage difference is: 0 volts

When this happens, both sides of the parallel bridge network are said to be balanced because the voltage at point C is the same value as the voltage at point D with their difference being zero.

Now let’s consider what would happen if we reversed the position of the two resistors, R3 and R4 in the second parallel branch with respect to R1 and R2.



With resistors, R3 and R4 reversed, the same current flows through the series combination and the voltage at point D, which is also the voltage drop across resistor, R4 will be:

VR4 = 0.4A × 10Ω = 4 volts

Now with VR4 having 4 volts dropped across it, the voltage difference between points C and D will be 4 volts as: C = 8 volts and D = 4 volts. Then the difference this time is: 8 – 4 = 4 volts

The result of swapping the two resistors is that both sides or “arms” of the parallel network are different as they produce different voltage drops. When this happens the parallel network is said to be unbalanced as the voltage at point C is at a different value to the voltage at point D.

Then we can see that the resistance ratio of these two parallel arms, ACB and ADB, results in a voltage difference between 0 volts (balanced) and the maximum supply voltage (unbalanced), and this is the basic principal of the Wheatstone Bridge Circuit.

So we can see that a Wheatstone bridge circuit can be used to compare an unknown resistance RX with others of a known value, for example, R1 and R2, have fixed values, and R3 could be variable. If we connected a voltmeter, ammeter or classically a galvanometer between points C and D, and then varied resistor, R3until the meters read zero, would result in the two arms being balanced and the value of RX, (substituting R4) known as shown.