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A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer is
- a)9
- b)90
- c)900
- d)990
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A wire potentiometer of length 11 m and resistance 1 /m balances a sta...
Current in AC: 1.081/10.18 = 0.1 A
Voltage across AB = 0.1 x 11 = 1.1V
Voltage across Rh = 2 - 1.1 = 0.9 V
Then, Rh = 0.9/0.1 = 9 Ω
Most Upvoted Answer
A wire potentiometer of length 11 m and resistance 1 /m balances a sta...
Given data:
- Length of wire potentiometer = 11 m
- Resistance of wire potentiometer = 1 ohm/m
- Standard cell voltage = 1.018 V at 10 m 18 cm
- Battery voltage = 2.0 V
Calculating resistance of wire potentiometer:
- Resistance of wire potentiometer = 1 ohm/m
- Length of wire potentiometer = 11 m
- Total resistance of wire potentiometer = 1 ohm/m x 11 m = 11 ohms
Calculating resistance at 10 m 18 cm:
- Total resistance at 10 m 18 cm = (10 + 18/100) x 1 = 10.18 ohms
Calculating resistance remaining:
- Remaining resistance = Total resistance - Resistance at 10 m 18 cm
- Remaining resistance = 11 ohms - 10.18 ohms = 0.82 ohms
Calculating series resistance:
- Battery voltage = 2.0 V
- Voltage drop across remaining resistance = Battery voltage - Standard cell voltage
- Voltage drop across remaining resistance = 2.0 V - 1.018 V = 0.982 V
- Series resistance = Voltage drop / Current
- Series resistance = 0.982 V / 1 A = 0.982 ohms
Therefore, the value of the series resistance connected to the potentiometer is 0.982 ohms, which is closest to option 'A' (9 ohms).
- Length of wire potentiometer = 11 m
- Resistance of wire potentiometer = 1 ohm/m
- Standard cell voltage = 1.018 V at 10 m 18 cm
- Battery voltage = 2.0 V
Calculating resistance of wire potentiometer:
- Resistance of wire potentiometer = 1 ohm/m
- Length of wire potentiometer = 11 m
- Total resistance of wire potentiometer = 1 ohm/m x 11 m = 11 ohms
Calculating resistance at 10 m 18 cm:
- Total resistance at 10 m 18 cm = (10 + 18/100) x 1 = 10.18 ohms
Calculating resistance remaining:
- Remaining resistance = Total resistance - Resistance at 10 m 18 cm
- Remaining resistance = 11 ohms - 10.18 ohms = 0.82 ohms
Calculating series resistance:
- Battery voltage = 2.0 V
- Voltage drop across remaining resistance = Battery voltage - Standard cell voltage
- Voltage drop across remaining resistance = 2.0 V - 1.018 V = 0.982 V
- Series resistance = Voltage drop / Current
- Series resistance = 0.982 V / 1 A = 0.982 ohms
Therefore, the value of the series resistance connected to the potentiometer is 0.982 ohms, which is closest to option 'A' (9 ohms).
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Community Answer
A wire potentiometer of length 11 m and resistance 1 /m balances a sta...
Current in AC: 1.081/10.18 = 0.1 A
Voltage across AB = 0.1 x 11 = 1.1V
Voltage across Rh = 2 - 1.1 = 0.9 V
Then, Rh = 0.9/0.1 = 9 Ω
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A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer isa)9b)90c)900d)990Correct answer is option 'A'. Can you explain this answer?
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A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer isa)9b)90c)900d)990Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer isa)9b)90c)900d)990Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer isa)9b)90c)900d)990Correct answer is option 'A'. Can you explain this answer?.
A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer isa)9b)90c)900d)990Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer isa)9b)90c)900d)990Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer isa)9b)90c)900d)990Correct answer is option 'A'. Can you explain this answer?.
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A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer isa)9b)90c)900d)990Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer isa)9b)90c)900d)990Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A wire potentiometer of length 11 m and resistance 1 /m balances a standard cell voltage of 1.018 V at a length of 10 m 18 cm. If the voltage of the battery supplying the current through the potentiometer is 2.0 V, then the value of the series resistance connected to the potentiometer isa)9b)90c)900d)990Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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