Home | | Physics 12th Std | Electric Cells and Batteries: Solved Example Problems

Chapter: 12th Physics : Current Electricity

Electric Cells and Batteries: Solved Example Problems

Physics : Current Electricity: Electric Cells and Batteries: Solved Example Problems


Determination of internal resistance: Solved Example Problems

EXAMPLE 2.17

A battery has an emf of 12 V and connected to a resistor of 3 Ω. The current in the circuit is 3.93 A. Calculate (a) terminal voltage and the internal resistance of the battery (b) power delivered by the battery and power delivered to the resistor

Solution

The given values I = 3.93 A, ξ = 12 V, R = 3 Ω

(a) The terminal voltage of the battery is equal to voltage drop across the resistor

V = IR = 3.93 × 3 = 11.79 V

The internal resistance of the battery,

r =  |ξ –V / V| R = | 12 −11 .79 /11 .79 |  × 3 = 0.05 Ω

The power delivered by the battery P = Iξ = 3.93 × 12 = 47.1 W

The power delivered to the resistor = I2 R = 46.3 W

The remaining power = (47.1 – 46.3) P = 0.772 W is delivered to the internal resistance and cannot be used to do useful work. (it is equal to I2 r).

 

Cells in series: Solved Example Problems


EXAMPLE 2.18

From the given circuit,


Find

i) Equivalent emf of the combination

ii) Equivalent internal resistance

iii) Total current

iv) Potential difference across external resistance

v) Potential difference across each cell

Solution

i) Equivalent emf of the combination ξeq = nξ = 4 9 = 36 V

ii) Equivalent internal resistance req = nr = 4 × 0.1 = 0.4 Ω

iii) Total current I = nξ / R +nr

 = [4 ×9] / 10 + ( 4 ×0.1)

= [4 ×9] /  [10 +0 .4] = 36 /10.4

I = 3.46 A

iv) Potential difference across external resistance V = IR = 3.46 × 10 = 34.6 V. The remaining 1.4 V is dropped across the internal resistance of cells.

v) Potential difference across each cell V/n = 34.6/4 = 8 .65V

 


Cells in parallel: Solved Example Problems



EXAMPLE 2.19

From the given circuit


Find

i) Equivalent emf

ii) Equivalent internal resistance

iii) Total current (I)

iv) Potential difference across each cell

v) Current from each cell

Solution

i) Equivalent emf ξeq = 5 V

ii) Equivalent internal resistance,

Req = r/n = 0 .5/4 = 0.125Ω

iii) total current, 


I ≈ 0.5 A

iv) Potential difference across each cell V = IR = 0.5 × 10 = 5 V

v) Current from each cell, I ′ = I/n

 I ′ = 0.5/4 = 0.125 A



Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Physics : Current Electricity : Electric Cells and Batteries: Solved Example Problems |


Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.