A potentiometer wire of length 1-0 m has a resistance of 15 - It is connected to a 5V battery in series with a resistance of 5 - Determine the emf of the primary cell which gives a balance point at 60 cm-

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Standard XII

Physics

Potentiometer

A potentiomet...

Question

A potentiometer wire of length 1.0 m has a resistance of 15 $Ω$ . It is connected to a 5V battery in series with a resistance of 5 $Ω$ . Determine the emf of the primary cell which gives a balance point at 60 cm.

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Solution

current flowing

$I = \frac{V}{R} = \frac{5}{5 + 15} = \frac{1}{4} A$

Resistance of 60 cm wire is $\frac{15}{100} \times 60 Ω = 9 Ω$

voltage drop on 60 cm wire is V=IR= $\frac{1}{4} \times 9 = 2.25 V$

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A potentiometer wire of length 1.0 m has a resistance of 15 Ω. It is connected to a 5V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.

current flowing I=VR=55+15=14A Resistance of 60 cm wire is 15100×60Ω=9Ω voltage drop on 60 cm wire is V=IR=14×9=2.25V

A potentiometer wire of length 1.0 m has a resistance of 15 $Ω$ . It is connected to a 5V battery in series with a resistance of 5 $Ω$ . Determine the emf of the primary cell which gives a balance point at 60 cm.

current flowing

$I = \frac{V}{R} = \frac{5}{5 + 15} = \frac{1}{4} A$

Resistance of 60 cm wire is $\frac{15}{100} \times 60 Ω = 9 Ω$

voltage drop on 60 cm wire is V=IR= $\frac{1}{4} \times 9 = 2.25 V$