Question

Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the Wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf 8 and the balance point found similarly, turns out to be at 82.3 cm length of the wire. What is the value ε.

Answer 1

$<b><i>Hope it will help you$

Answer 2

it is given that,

constant emf of the given standard cell, E1 = 1.02V

balance point on the wire, l1 = 67.3cm

an unknown cell of emf Ωμ, replaced the standard cell. due to this new balance point on the wire, l2 = 82.3cm

now use the relation connecting emf and balance point,

E1/l1 =ε/l2

or, ε = E1 (l2/l1)

= 1.02 × (82.3/67.3)

= 1.02 × 1.22288262

= 1.24734027 ≈ 1.247V

hence, value of ε = 1.247 V