Question

Question 3.11: A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Class 12 - Physics - Current Electricity Current Electricity Page-128

Answer 1

During charging , the electric current is sent into 8V battery.
let E = 8V, E'= 120V , r = 0.5 ohm and R = 15.5 ohm
apply Kirchoff's law,
E - Ir - IR + E' = 0
=> I = (E' - E)/(r + R)
=> I = (120 - 8)/(0.5 + 15.5)
=> I = 112/16 = 7A

now, terminal voltage of the battery during charging, V = E + Ir
V = 8 + 7 × 0.5 = 8 + 3.5 = 11.5 V

A series resistance is joined in the charging circuit to limit the excessive current so that charging is slow and permanent.

Answer 2

Hey !!

(i) 3ε = 6 ; ε = 2

(ii) For power to be maximum in the circuit the load resistance will be equal to internal resistance of circuit.

  Let r is the internal resistance of one cell then,

          V = 3E + 3Ir

when, V = 0.1 = 2A , so r = E/I

                          = 1Ω

So for maximum power dissipation ( ∴ r = R )

R = 3R = 3Ω

For maximum power dissipation  

  I = 3E / R + 3R

    = 3 × 2 / 3 + 3

FINAL RESULT = 11.5 volt

HOPE IT HELPS YOU !!