How to find the Equivalent Resistance of an Unbalanced Wheatstone Bridge?
There are many ways to find out the equivalent resistance of an unbalanced Wheatstone Bridge.
Method 1
By Using Kirchhoff's Laws
Let us consider an example :
Example 1
Find the equivalent resistance between the points A and B
Solution :
Applying Kirchhoff's Loop Rule in loop 1
5I1+10I2-10(I-I1)=0
5I1+10I2-10I+10I1=0
15I1+10I2-10I=0
5[3I1+2I2-2I]=0
3I1+2I2-2I=0
3I1+2I2=2I — (1)
5I1+10I2-10I+10I1=0
15I1+10I2-10I=0
5[3I1+2I2-2I]=0
3I1+2I2-2I=0
3I1+2I2=2I — (1)
Again, Applying Kirchhoff's Loop Rule in loop 2
10I2+5(I-I1+I2)-10(I1-I2)=0
10I2+5I-5I1+5I2-10I1+10I2=0
-15I1+25I2+5I=0
5[-3I1+5I2+I]=0
-3I1+5I2+I=0
-3I1+5I2=-I — (2)
10I2+5I-5I1+5I2-10I1+10I2=0
-15I1+25I2+5I=0
5[-3I1+5I2+I]=0
-3I1+5I2+I=0
-3I1+5I2=-I — (2)
3I1+2I2=2I
-3I1+5I2=-I
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0 + 7I2 = I
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0 + 7I2 = I
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7I2 = I
I2 = I/7
Putting I2 in equation (1)
3I1 + 2(I/7) = 2I
3I1 = 2I-2I/7
3I1 = (14I-2I)7
3I1 = 12I/7
I1 = 4I/7
3I1 = 2I-2I/7
3I1 = (14I-2I)7
3I1 = 12I/7
I1 = 4I/7
Once again, applying Kirchhoff's Loop Rule in loop 3
10(I-I1)+5(I-I1+I2)=V
10I-10I1+5I-5I1+5I2=V
15I-15I1+5I2 = V
15I-15(4I/7)+5(I/7)=V [Putting the values of I1 and I2]
15I-(60I/7)+(5I/7)=V
(105I-60I+5I)/7=V
(105I-60I+5I)/7=V
50I/7 = V
V = 50I/7
Now, Using Ohm's Law
Now, Using Ohm's Law
V=IReq
Req = V/I
Req = V/I
Req = (50I/7)/I
Req = 50/7
Req = 7.14 Ω (approx)
Req = 50/7
Req = 7.14 Ω (approx)
Example 2
Find the equivalent resistance between the points A and B
Solution :
Applying Kirchhoff's Loop Rule in loop 1
5I1+5I2-10(I-I1)=0
5I1+5I2-10I+10I1=0
15I1+5I2-10I=0
5[3I1+I2-2I]=0
3I1+I2-2I=0
3I1+I2=2I — (1)
5I1+5I2-10I+10I1=0
15I1+5I2-10I=0
5[3I1+I2-2I]=0
3I1+I2-2I=0
3I1+I2=2I — (1)
Again, Applying Kirchhoff's Loop Rule in loop 2
5I2+5(I-I1+I2)-10(I1-I2)=0
5I2+5I-5I1+5I2-10I1+10I2=0
-15I1+20I2+5I=0
5[-3I1+4I2+I]=0
-3I1+4I2+I=0
-3I1+4I2=-I — (2)
5I2+5I-5I1+5I2-10I1+10I2=0
-15I1+20I2+5I=0
5[-3I1+4I2+I]=0
-3I1+4I2+I=0
-3I1+4I2=-I — (2)
3I1+I2=2I
-3I1+4I=-I
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0 + 5I2 = I
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0 + 5I2 = I
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5I2 = I
I2 = I/5
Putting I2 in equation (1)
3I1 + (I/5) = 2I
3I1 = 2I-I/5
3I1 = (10I-I)5
3I1 = 9I/5
I1 = 3I/5
3I1 = 2I-I/5
3I1 = (10I-I)5
3I1 = 9I/5
I1 = 3I/5
Once again, applying Kirchhoff's Loop Rule in loop 3
10(I-I1)+5(I-I1+I2)=V
10I-10I1+5I-5I1+5I2=V
15I-15I1+5I2 = V
15I-15(3I/5)+5(I/5)=V [Putting the values of I1 and I2]
15I-(45I/5)+(5I/5)=V
(75I-45I+5I)/7=V
(75I-45I+5I)/7=V
35I/7 = V
V = 35I/7
V=5I
Now, Using Ohm's Law
V=5I
Now, Using Ohm's Law
V=IReq
Req = V/I
Req = V/I
Req = (5I)/I
Req = 5 Ω
Req = 5 Ω
Method 2
By Using Delta to Star Conversion
Delta to Star Conversion Formula
R1=RBRC/(RA+RB+RC)
R2=RARC/(RA+RB+RC)
R3=RARB/(RA+RB+RC)
Star to Delta Conversion Formula
RA=(R1R2+R2R3+R3R1)/R1
RB=(R1R2+R2R3+R3R1)/R2
RC=(R1R2+R2R3+R3R1)/R3
Example 1
Find the equivalent resistance between the points A and B
Solution :
Coming Soon
