Evaluating the Wheatstone Bridge
Spring 2010
Physics 338
Dr. Adam T. Whitten
To find the Thévenin equivalent circuit for the Wheatstone bridge pictured
to the right requires finding VTH when terminals a and b are unconnected
and ITH when terminals a and b are shorted together. For VTH we have
voltage dividers on the left and right sides that give us Va and Vb .
Va = (12 V)
1.0k
3.3k
2.2k
1.0k
1.5k
3.3k
2.2k
3.3 k
2.2 k
= 9.21 V
Vb = (12 V)
= 7.14 V
1.0 k + 3.3 k
1.5 k + 2.2 k
VTH = Va − Vb = 2.07 V
For ITH the majority of the current follows the “path of least resistance”
indicated by the bold arrows in the picture to the right. However, there
is still a non-zero current through the 1.5 k and 3.3 k resistors,
because there is a potential difference across these resistors.
+12V
To find ITH redraw the resistors as in the figure to the right. Since the terminals are
shorted together we have a parallel combination of resistors in series with another parallel
combination of resistors. The top parallel combination gives a 600 Ω effective resistance
and the bottom parallel combination gives a 1.32 kΩ effective resistance for a total
efective resistance of 1.92 kΩ.
ITH =
1.5k
a b
+12V
12 V
= 6.25 mA
1.92 kΩ
=⇒
RTH =
VTH
2.07 V
==
= 331 Ω
ITH
6.25 mA
This means that the voltage drop across the upper parallel combination of resistors is:
Vu = ITH (600 Ω) = 3.75 V
and the voltage drop across the lower parallel combination of resistors is:
Vl = ITH (1.32 kΩ) = 8.25 V
The currents (labeled IR ) are then:
I1.0k =
3.75 V
= 3.75 mA
1.0 kΩ
I3.3k =
8.25 V
= 2.5 mA
3.3 kΩ
I1.5k =
I2.2k =
1
3.75 V
= 2.5 mA
1.5 kΩ
8.25 V
= 3.75 mA
2.2 kΩ
1.0k
1.5k
3.3k
2.2k