Lab 5: Wheatstone Bridge
Only 7 more labs to go!!
The Wheatstone Bridge is a commonly used circuit for making electrical measurements.
When will the current through the ammeter
be equal to zero?
This will happen if we choose the resistor to
be just right!
If we assume no current flow through the
ammeter then:
Vb  VD  I b Rb  I D RD
Ia
Ra
IC
RC
A
Ib
Rb
ID
RD
Va  VC  I a Ra  I C RC
Since no current flows through the ammeter:
I a  I b and I C  I D
This means we can divide the above equations and get this ratio:
Ra RC
RD

 RC  Ra
Rb RD
Rb
Remember that:
L
R
A
Where  is the resistivity, L is the length and A is the
cross sectional area.
Using this relationship we can rewrite:
LD

RD
LD
A
RC  Ra
 Ra
 Ra
Lb
Rb
L
b

A
Metal Bar
Ra
RC
A
Resistance Wire
Lb
LD
LD
RC  Ra
Lb
R1= 10 
What is the equivalent resistance?
First reduce the parallel array to one resistor.
12 V
R4= 45 
R3= 30 
R2= 20 
1
1
1
1

 
R234 R2 R3 R4
1
1
1
1 964

 

R234 20 30 45
180
R234 
180
 9.5 
19
R1= 10 
12 V
12 V
R234= 9.5 
R1234 = 19.5 
I = 0.62 A
What is the current through R1 ?
V
12V
I 
 0.62 A
R 19.5
R1234 = 19.5 
12 V
I = 0.62 A
What is the voltage dropped by R1?
V1  IR  0.62 A 10  6.2 V
So this means that the voltage dropped
by the parallel resistors is:
R1= 10 
6.2 V
R234= 9.5 
5.8 V
12 V
V234  12V  6.2V  5.8V
Can you find the current through each resistor? Use Ohm’s Law
V
I
R