A potentiometer wire of length 1m has a resistance of 10- It is connected to a 6V battery in series with a resistance of 5- Determine the emf of the primary cell which gives a balance point at 40cm-

Voltage across potentiometer = I R = 6 15 × 10 =4V voltage across 1m wire = 4VPotential gradient = 4 1 =4V/m=K EMF = Kl ( l = balancing ...

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A potentiomet...

Question

A potentiometer wire of length $1 m$ has a resistance of $10 Ω$ . It is connected to a $6 V$ battery in series with a resistance of $5 Ω$ . Determine the emf of the primary cell which gives a balance point at $40 c m$ .

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Solution

Voltage across potentiometer $= I R = \frac{6}{15} \times 10 = 4 V$
voltage across 1m wire = 4V
Potential gradient $= \frac{4}{1} = 4 V / m = K$
EMF = Kl ( l = balancing point)
E $= 4 \times \frac{40}{100}$
$ε$ =1.6 V

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A potentiometer wire of length 1m has a resistance of 10Ω. It is connected to a 6V battery in series with a resistance of 5Ω. Determine the emf of the primary cell which gives a balance point at 40cm.

Voltage across potentiometer =IR=615×10=4Vvoltage across 1m wire = 4VPotential gradient =41=4V/m=KEMF = Kl ( l = balancing point)E =4×40100ε =1.6 V

A potentiometer wire of length $1 m$ has a resistance of $10 Ω$ . It is connected to a $6 V$ battery in series with a resistance of $5 Ω$ . Determine the emf of the primary cell which gives a balance point at $40 c m$ .

Voltage across potentiometer $= I R = \frac{6}{15} \times 10 = 4 V$
voltage across 1m wire = 4V
Potential gradient $= \frac{4}{1} = 4 V / m = K$
EMF = Kl ( l = balancing point)
E $= 4 \times \frac{40}{100}$
$ε$ =1.6 V