Q. 194.3( 3 Votes )

# The internal resi

Answer :

a)0.3A, b)3A

Given,

The emf of the battery, E=6V

The internal resistance of the battery, r_{1} when discharged=10Ω

The internal resistance of the battery, r_{2} when charged=1Ω

Potential difference provided by the charger, E_{c}=9V

Formula used:

When the accumulator battery is connected to a charger, the current through the internal resistance depends on the net emf available across the resistor.

Hence Net emf across the resistor in the case of charging will be the difference between the provided potential difference and the potential difference rating of the battery.

Solution:

a) When the battery is being charged, the net emf, E_{net} across the resistance, r_{1} will be,

Or,

Hence from the relation, , we can find the current, i by substituting r_{1} as 10 Ω,

Or,

b) When the battery is completely charged, the internal resistance r_{2} will be 1Ω. The net emf across the resistance will be the same, and is 3V.

Hence from the relation, we can find the current, i through the resistance by substituting r_{2} as 1Ω,

Or,

Hence the current through the internal resistance while charging and after completely charged are 0.3A and 3A.

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