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In a Wheatstone's bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is
- a)R/2
- b)R
- c)2 R
- d)R/4
Correct answer is option 'B'. Can you explain this answer?
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In a Wheatstones bridge all the four arms have equal resistance R. If ...
Solution:
A Wheatstone Bridge is a circuit used to compare an unknown resistance with a known resistance. It consists of four resistors R1, R2, R3 and R4 connected between a voltage source and a galvanometer.
Given, R1 = R2 = R3 = R4 = R and Rg = R.
To find out the equivalent resistance of the combination, we can use the following steps:
Step 1: Apply Kirchhoff's first law to the junction point A.
I1 + I3 = I2 + I4
Step 2: Apply Ohm's law to each resistor.
I1 = V/R
I2 = (V - IR)/R
I3 = IR/R
I4 = (V - IR)/R
Step 3: Substitute the values of I1, I2, I3 and I4 in the equation obtained in step 1.
V/R + IR/R = (V - IR)/R + (V - IR)/R
Step 4: Solve for I.
I = V/2R
Step 5: Apply Kirchhoff's second law to the loop ABCDA.
IR - IR + I*Rg - V = 0
Step 6: Solve for V.
V = IRg
Step 7: Substitute the value of V in the equation obtained in step 4.
I = IRg/2R
Step 8: Simplify.
I = Ig/2
Step 9: Apply Ohm's law to the equivalent resistance.
Req = V/I
Req = 2R
Therefore, the equivalent resistance of the combination as seen by the battery is 2R.
Hence, option B is the correct answer.
A Wheatstone Bridge is a circuit used to compare an unknown resistance with a known resistance. It consists of four resistors R1, R2, R3 and R4 connected between a voltage source and a galvanometer.
Given, R1 = R2 = R3 = R4 = R and Rg = R.
To find out the equivalent resistance of the combination, we can use the following steps:
Step 1: Apply Kirchhoff's first law to the junction point A.
I1 + I3 = I2 + I4
Step 2: Apply Ohm's law to each resistor.
I1 = V/R
I2 = (V - IR)/R
I3 = IR/R
I4 = (V - IR)/R
Step 3: Substitute the values of I1, I2, I3 and I4 in the equation obtained in step 1.
V/R + IR/R = (V - IR)/R + (V - IR)/R
Step 4: Solve for I.
I = V/2R
Step 5: Apply Kirchhoff's second law to the loop ABCDA.
IR - IR + I*Rg - V = 0
Step 6: Solve for V.
V = IRg
Step 7: Substitute the value of V in the equation obtained in step 4.
I = IRg/2R
Step 8: Simplify.
I = Ig/2
Step 9: Apply Ohm's law to the equivalent resistance.
Req = V/I
Req = 2R
Therefore, the equivalent resistance of the combination as seen by the battery is 2R.
Hence, option B is the correct answer.
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In a Wheatstones bridge all the four arms have equal resistance R. If ...
R ,R through galvanometer would shorcircuit ,and above and below resistance are in series hence add up to 2R each and the resultant are parallel hence R
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In a Wheatstones bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery isa)R/2b)Rc)2 Rd)R/4Correct answer is option 'B'. Can you explain this answer?
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