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Standard XII

Physics

Potentiometer

A potentiomet...

Question

A potentiometer wire of length $200$ cm has a resistance of $20 Ω$ . It is connected in series with a resistance of $10 Ω$ and an accumulator of emf $6$ V having negligible internal resistance. A source of $2.4$ V is balanced against a length L of the potentiometer wire. The value of L is?

100

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120

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110

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140

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Solution

The correct option is B $120$ cm
$I = \frac{V}{R} = \frac{6}{10 + 20} = \frac{6}{30} = \frac{1}{5} A$
Voltage across wire $= 6 - \frac{1}{5} \times 10 = 4 V$
Voltage per unit length $= \frac{4}{200} = 0.02 v / s$
Length for balancing $= \frac{2.4}{2} \times 100 c m = 120 c m$
Answer B

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A potentiometer wire of length 200 cm has a resistance of $20 Ω$ . It is connected in series with a resistance of $10 Ω$ and an accumulator of emf 6 V having negligible internal resistance. A source of 2.4 V is balanced against a length L of the potentiometer wire. The value of L is

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A potentiometer wire of length 200cm has a resistance of 20Ω. It is connected in series with a resistance of 10Ω and an accumulator of emf 6V having negligible internal resistance. A source of 2.4V is balanced against a length L of the potentiometer wire. The value of L is?

The correct option is B 120cmI=VR=610+20=630=15AVoltage across wire =6−15×10=4VVoltage per unit length =4200=0.02v/sLength for balancing =2.42×100cm=120cmAnswer B

A potentiometer wire of length $200$ cm has a resistance of $20 Ω$ . It is connected in series with a resistance of $10 Ω$ and an accumulator of emf $6$ V having negligible internal resistance. A source of $2.4$ V is balanced against a length L of the potentiometer wire. The value of L is?

The correct option is B $120$ cm
$I = \frac{V}{R} = \frac{6}{10 + 20} = \frac{6}{30} = \frac{1}{5} A$
Voltage across wire $= 6 - \frac{1}{5} \times 10 = 4 V$
Voltage per unit length $= \frac{4}{200} = 0.02 v / s$
Length for balancing $= \frac{2.4}{2} \times 100 c m = 120 c m$
Answer B