DC Bridges

DC Bridges

DC bridges can be operated with only DC voltage signal. DC bridges are useful for measuring the value of unknown resistance, which is present in the bridge. Wheatstone’s Bridge is an example of DC bridge.

Now, let us discuss about Wheatstone’s Bridge in order to find the unknown resistance’s value.

Wheatstone’s Bridge

Wheatstone’s bridge is a simple DC bridge, which is mainly having four arms. These four arms form a rhombus or square shape and each arm consists of one resistor.

To find the value of unknown resistance, we need the galvanometer and DC voltage source. Hence, one of these two are placed in one diagonal of Wheatstone’s bridge and the other one is placed in another diagonal of Wheatstone’s bridge.

Wheatstone’s bridge is used to measure the value of medium resistance. The circuit diagram of Wheatstone’s bridge is shown in below figure.

Wheatstone’s bridge

In above circuit, the arms AB, BC, CD and DA together form a rhombus or square shape. They consist of resistors R2R2R4R4R3R3 and R1R1 respectively. Let the current flowing through these resistor arms is I2I2I4I4I3I3 and I1I1 respectively and the directions of these currents are shown in the figure.

The diagonal arms DB and AC consists of galvanometer and DC voltage source of V volts respectively. Here, the resistor, R3R3 is a standard variable resistor and the resistor, R4R4 is an unknown resistor. We can balance the bridge, by varying the resistance value of resistor, R3R3.

The above bridge circuit is balanced when no current flows through the diagonal arm, DB. That means, there is no deflection in the galvanometer, when the bridge is balanced.

The bridge will be balanced, when the following two conditions are satisfied.

  • The voltage across arm AD is equal to the voltage across arm AB. i.e.,
VAD=VABVAD=VAB

I1R1=I2R2⇒I1R1=I2R2Equation 1

  • The voltage across arm DC is equal to the voltage across arm BC. i.e.,
VDC=VBCVDC=VBC

I3R3=I4R4⇒I3R3=I4R4Equation 2

From above two balancing conditions, we will get the following two conclusions.

  • The current flowing through the arm AD will be equal to that of arm DC. i.e.,
I1=I3I1=I3
  • The current flowing through the arm AB will be equal to that of arm BC. i.e.,
I2=I4I2=I4

Take the ratio of Equation 1 and Equation 2.

I1R1I3R3=I2R2I4R4I1R1I3R3=I2R2I4R4Equation 3

Substitute, I1=I3I1=I3 and I2=I4I2=I4 in Equation 3.

I3R1I3R3=I4R2I4R4I3R1I3R3=I4R2I4R4
R1R3=R2R4⇒R1R3=R2R4
R4=R2R3R1⇒R4=R2R3R1

By substituting the known values of resistors R1R1R2R2 and R3R3 in above equation, we will get the value of resistor,R4R4.