# Answer to Question #33261 in Electric Circuits for Frank Loyld

Question #33261

A Wheatstone bridge ABCD (labeled in clockwise direction) with a galvanometer G joined to the junctions B and D. A cell E is connected across AC. Resistances of 15, 10, 4 and X ohms are inserted into the arms AB, BC, CD and DA respectively. The value of X is _________ ohms.

A. 2.67

B. 1.34

C. 3.22

D. 2.13

A. 2.67

B. 1.34

C. 3.22

D. 2.13

Expert's answer

R1=15 ohms

R2= 10 ohms

R3= 4 ohms

First, Kirchhoff's first rule is used to find the currents in junctions B and D:

B) I1 - I2 - Ig = 0 ( Ig - current through the galvanometer)

D) Ix - I3 - Ig = 0 ( Ix - current trough Rx)

Then, Kirchhoff's second rule is used for finding the voltage in the loops ABD and BCD:

ABD) I1*R1 + Ix*Rx + Ig*Rg = 0

BCD) I2*R2 + I3*R3+ Ig*Rg = 0

When the bridge is balanced, then Ig = 0, so the second set of equations can be rewritten as:

I1*R1 = - Ix*Rx

I2*R2 = - I3*R3

and

I1=I2

Ix=I3

From the first rule and the second rule we get:

Rx=(R3*R2)/R1

Answer: A, 2,67 ohms

R2= 10 ohms

R3= 4 ohms

First, Kirchhoff's first rule is used to find the currents in junctions B and D:

B) I1 - I2 - Ig = 0 ( Ig - current through the galvanometer)

D) Ix - I3 - Ig = 0 ( Ix - current trough Rx)

Then, Kirchhoff's second rule is used for finding the voltage in the loops ABD and BCD:

ABD) I1*R1 + Ix*Rx + Ig*Rg = 0

BCD) I2*R2 + I3*R3+ Ig*Rg = 0

When the bridge is balanced, then Ig = 0, so the second set of equations can be rewritten as:

I1*R1 = - Ix*Rx

I2*R2 = - I3*R3

and

I1=I2

Ix=I3

From the first rule and the second rule we get:

Rx=(R3*R2)/R1

Answer: A, 2,67 ohms

## Comments

## Leave a comment