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A wire potentiometer of length 11 m and resistance 1 Ω/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 (2004)
- a)9 Ω
- b)90 Ω
- c)900 Ω
- d)990 Ω
Correct answer is option 'A'. Can you explain this answer?
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A wire potentiometer of length 11 m and resistance 1 Ω/m balances a s...
Voltage cross Rh =2 - 1.1 = 0.9 V
Then, 
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A wire potentiometer of length 11 m and resistance 1 Ω/m balances a s...
Given data:
- Length of wire potentiometer (L) = 11 m
- Resistance of wire potentiometer (R) = 1 Ω/m
- Voltage of standard cell (V_s) = 1.018 V
- Length at which standard cell balances the voltage (L_b) = 10 m 18 cm
- Voltage of battery supplying the current (V_b) = 2.0 V
To find:
- Value of the series resistance connected to the potentiometer
Finding the potential gradient:
- The potential gradient (G) is defined as the change in potential per unit length of the wire potentiometer.
- It can be calculated using the formula: G = V_s / L, where V_s is the voltage of the standard cell and L is the length of the wire potentiometer.
- Substituting the given values, G = 1.018 V / 11 m = 0.0925 V/m
Finding the balancing length:
- The balancing length (L_b) is the length at which the voltage of the battery supplying the current balances the voltage of the standard cell.
- It can be calculated using the formula: L_b = V_b / G, where V_b is the voltage of the battery supplying the current and G is the potential gradient.
- Substituting the given values, L_b = 2.0 V / 0.0925 V/m = 21.62 m
Calculating the resistance of the wire:
- The resistance of the wire (R_w) is equal to the product of the length (L_b) at which balancing occurs and the resistance per unit length (R) of the wire potentiometer.
- Substituting the given values, R_w = 21.62 m * 1 Ω/m = 21.62 Ω
Calculating the value of the series resistance:
- The series resistance (R_s) connected to the potentiometer can be calculated using the formula: R_s = R_w - R, where R_w is the resistance of the wire and R is the resistance per unit length.
- Substituting the given values, R_s = 21.62 Ω - 1 Ω/m = 20.62 Ω
Therefore, the value of the series resistance connected to the potentiometer is 20.62 Ω, which is closest to option A: 9 Ω.
- Length of wire potentiometer (L) = 11 m
- Resistance of wire potentiometer (R) = 1 Ω/m
- Voltage of standard cell (V_s) = 1.018 V
- Length at which standard cell balances the voltage (L_b) = 10 m 18 cm
- Voltage of battery supplying the current (V_b) = 2.0 V
To find:
- Value of the series resistance connected to the potentiometer
Finding the potential gradient:
- The potential gradient (G) is defined as the change in potential per unit length of the wire potentiometer.
- It can be calculated using the formula: G = V_s / L, where V_s is the voltage of the standard cell and L is the length of the wire potentiometer.
- Substituting the given values, G = 1.018 V / 11 m = 0.0925 V/m
Finding the balancing length:
- The balancing length (L_b) is the length at which the voltage of the battery supplying the current balances the voltage of the standard cell.
- It can be calculated using the formula: L_b = V_b / G, where V_b is the voltage of the battery supplying the current and G is the potential gradient.
- Substituting the given values, L_b = 2.0 V / 0.0925 V/m = 21.62 m
Calculating the resistance of the wire:
- The resistance of the wire (R_w) is equal to the product of the length (L_b) at which balancing occurs and the resistance per unit length (R) of the wire potentiometer.
- Substituting the given values, R_w = 21.62 m * 1 Ω/m = 21.62 Ω
Calculating the value of the series resistance:
- The series resistance (R_s) connected to the potentiometer can be calculated using the formula: R_s = R_w - R, where R_w is the resistance of the wire and R is the resistance per unit length.
- Substituting the given values, R_s = 21.62 Ω - 1 Ω/m = 20.62 Ω
Therefore, the value of the series resistance connected to the potentiometer is 20.62 Ω, which is closest to option A: 9 Ω.
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A wire potentiometer of length 11 m and resistance 1 Ω/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 (2004)a)9 Ωb)90 Ωc)900 Ωd)990 ΩCorrect answer is option 'A'. Can you explain this answer?
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A wire potentiometer of length 11 m and resistance 1 Ω/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 (2004)a)9 Ωb)90 Ωc)900 Ωd)990 ΩCorrect 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 Ω/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 (2004)a)9 Ωb)90 Ωc)900 Ωd)990 ΩCorrect 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 Ω/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 (2004)a)9 Ωb)90 Ωc)900 Ωd)990 ΩCorrect answer is option 'A'. Can you explain this answer?.
A wire potentiometer of length 11 m and resistance 1 Ω/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 (2004)a)9 Ωb)90 Ωc)900 Ωd)990 ΩCorrect 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 Ω/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 (2004)a)9 Ωb)90 Ωc)900 Ωd)990 ΩCorrect 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 Ω/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 (2004)a)9 Ωb)90 Ωc)900 Ωd)990 ΩCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A wire potentiometer of length 11 m and resistance 1 Ω/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 (2004)a)9 Ωb)90 Ωc)900 Ωd)990 ΩCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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