Sample Questions on Wheatstone Bridge
Question 1: Find the equivalent resistance between points A and C in the circuit shown in the figure below:
Answer:
The equivalent circuit of the circuit shown in the above figure is given as:
Since, 2/3 = 4/6
This is the circuit of a balanced Wheatstone bridge.
In the balanced state, VB = VD (where V represents potential)
So no current will flow through the 5 Ω resistance.
Now the equivalent resistance of sides AB and BC is R’ = 2 + 3 = 5 Ω.
The equivalent resistance of AD and DC arm R” = 4 + 6 = 10 Ω
If the equivalent resistance between the points A and C is R, and R is parallel combination of resistance R’ and R”,
⇒ 1 / R = (1 / R’) + (1 / R”)
⇒ 1/R = (1/5) + (1/10)
⇒ 1/R = (2 + 1) / 10
⇒ R = 10/3
⇒ R = 3.33 Ω
Question 2: The electric circuit of a balanced Wheatstone bridge is shown in Figure. Calculate the resistance x.
Answer:
Let the total resistance in arm BC be R. Since the bridge is balanced, therefore:
15/R = 5/10
⇒ R = (15 × 10)/5 = 30 Ω
Now, as R is parallel combination of 60 Ω and X.
⇒ 1/R = (1/X) + (1/60)
⇒ 1/30 = 1/X + (1/60)
⇒ 1/X = 1/30 – 1/60
⇒ 1/X = (2 – 1)/60 = 1 / 60
⇒ X = 60 Ω
Question 3: What is a Meter Bridge and what kind of precautions do we need to perform measurements using a Meter Bridge?
Answer:
A Meter bridge is a device based on the principle of the Wheatstone bridge, with the help of which the resistance and specific resistance of a conductor can be determined. In this, a 1-meter long wire acts as the proportional side.
Precautions while using Meter Bridge are:
- The ends of all the connection wires should be cleaned with sandpaper.
- The current should not flow in the circuit for a long time otherwise, its resistance increases due to the heating of the bridge wire. Therefore, the key in the cell circuit should be plugged in only when observations are to be made.
- The jockey should not be run by rubbing it on the meter bridge wire otherwise, the thickness of the wire will not remain the same at all places.
- A shunt with a galvanometer should be used initially while adjusting, but the shunt should be removed near the position of zero deflection.
- Only such a resistance plug should be removed from the resistance box so that the position of zero deflection is approximately in the middle of the bridge wire. In this case, the sensitivity of the bridge is maximum and the percentage error is minimum.
- All other plugs in the resistance box, except those that have been removed, should be tightly packed.
Question 4: In the following figure, find the current through the 4Ω resistor.
Answer:
Since, Q / P = S / R
⇒ 4 / 20 =10 / 50
It is an example of balanced wheat-stone bridge.
So, No current will flow through 16 Ω resistance.
As we know that current divide in inverse ratio, current through 4Ω resistance is,
= 1.4 × (20+4)+(50+10) / (50+10)
= 1.4 × (7/5)
= 1 A
Wheatstone Bridge
Wheatstone bridge is a device that is used to find the resistance of a conductor, in 1842, scientist Wheatstone proposed a theory, which is called the principle of Wheatstone bridge after his name. we can prove or establish the formula for Wheatstone by using Kirchhoff laws. Wheatstone bridge is simply an electric circuit used to measure an unknown electric resistance by balancing two-point of a bridge. Let’s get started!
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