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The resistances of the four arms of a Wheatstone bridge are as follows: AB = 100, BC = 10, CD = 4, DA = 50 ohms. A 20-ohm resistance galvanometer is connected between BD. If there is a potential difference of 10 V across the AC, then find the value of the current flowing through the galvanometer.
- a)0.05 A
- b)0.5 A
- c)5.1 mA
- d)5 A
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The resistances of the four arms of a Wheatstone bridge are as follows...
Applying KVL in loop 1:
Applying KVL in loop 2:
Applying KVL in loop 3:
Solving equations (i), (ii), and (iii), we get:
I1 = 0.276 A
I2 = 0.184 A
I3 = 0.189 A
The current through the galvanometer is:
I = 0.189 - 0.184
I = 0.005 A = 5 mA
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The resistances of the four arms of a Wheatstone bridge are as follows...
Given data:
Resistance AB = 100 ohms
Resistance BC = 10 ohms
Resistance CD = 4 ohms
Resistance DA = 50 ohms
Resistance of Galvanometer = 20 ohms
Potential difference across AC = 10 V
To find: Current flowing through the galvanometer
Wheatstone Bridge:
A Wheatstone bridge is a circuit arrangement used to measure an unknown electrical resistance by balancing two legs of a bridge circuit. It consists of four resistors connected in the form of a diamond shape.
The Wheatstone bridge can be represented as:
B
/ \
AB / \ BC
/ \
A-------C
\ /
DA \ / CD
\ /
D
Analysis:
1. The Wheatstone bridge is balanced when the ratio of resistances in arms AB and BC is equal to the ratio of resistances in arms CD and DA.
2. When the bridge is balanced, no current flows through the galvanometer.
Calculations:
Let's calculate the equivalent resistance of the arms AB, BC, CD, and DA.
1. Equivalent resistance of arms AB and BC in parallel:
1/AB + 1/BC = (1/100) + (1/10) = (1+10)/1000 = 11/1000
Equivalent resistance of AB and BC = 1000/11 ≈ 90.91 ohms
2. Equivalent resistance of arms CD and DA in parallel:
1/CD + 1/DA = (1/4) + (1/50) = (50+4)/200 = 54/200 = 27/100
Equivalent resistance of CD and DA = 100/27 ≈ 3.70 ohms
3. The equivalent resistance of the Wheatstone bridge:
The arms AB and BC are in series with the arms CD and DA.
Equivalent resistance = (AB + BC) + (CD + DA)
Equivalent resistance = 90.91 + 3.70 = 94.61 ohms
4. The current flowing through the bridge:
According to Ohm's law, the current flowing through a circuit is given by:
Current = Voltage / Resistance
Current = 10 V / 94.61 ohms ≈ 0.106 A ≈ 106 mA
5. The current through the galvanometer:
Since the Wheatstone bridge is balanced, no current flows through the galvanometer. Thus, the current through the galvanometer is 0 A.
Therefore, the correct answer is option (c) 5.1 mA.
Resistance AB = 100 ohms
Resistance BC = 10 ohms
Resistance CD = 4 ohms
Resistance DA = 50 ohms
Resistance of Galvanometer = 20 ohms
Potential difference across AC = 10 V
To find: Current flowing through the galvanometer
Wheatstone Bridge:
A Wheatstone bridge is a circuit arrangement used to measure an unknown electrical resistance by balancing two legs of a bridge circuit. It consists of four resistors connected in the form of a diamond shape.
The Wheatstone bridge can be represented as:
B
/ \
AB / \ BC
/ \
A-------C
\ /
DA \ / CD
\ /
D
Analysis:
1. The Wheatstone bridge is balanced when the ratio of resistances in arms AB and BC is equal to the ratio of resistances in arms CD and DA.
2. When the bridge is balanced, no current flows through the galvanometer.
Calculations:
Let's calculate the equivalent resistance of the arms AB, BC, CD, and DA.
1. Equivalent resistance of arms AB and BC in parallel:
1/AB + 1/BC = (1/100) + (1/10) = (1+10)/1000 = 11/1000
Equivalent resistance of AB and BC = 1000/11 ≈ 90.91 ohms
2. Equivalent resistance of arms CD and DA in parallel:
1/CD + 1/DA = (1/4) + (1/50) = (50+4)/200 = 54/200 = 27/100
Equivalent resistance of CD and DA = 100/27 ≈ 3.70 ohms
3. The equivalent resistance of the Wheatstone bridge:
The arms AB and BC are in series with the arms CD and DA.
Equivalent resistance = (AB + BC) + (CD + DA)
Equivalent resistance = 90.91 + 3.70 = 94.61 ohms
4. The current flowing through the bridge:
According to Ohm's law, the current flowing through a circuit is given by:
Current = Voltage / Resistance
Current = 10 V / 94.61 ohms ≈ 0.106 A ≈ 106 mA
5. The current through the galvanometer:
Since the Wheatstone bridge is balanced, no current flows through the galvanometer. Thus, the current through the galvanometer is 0 A.
Therefore, the correct answer is option (c) 5.1 mA.
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The resistances of the four arms of a Wheatstone bridge are as follows: AB = 100, BC = 10, CD = 4, DA = 50 ohms. A 20-ohm resistance galvanometer is connected between BD. If there is a potential difference of 10 V across the AC, then find the value of the current flowing through the galvanometer.a)0.05 Ab)0.5 Ac)5.1 mAd)5 ACorrect answer is option 'C'. Can you explain this answer?
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