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arsen [322]
3 years ago
14

What is the emf of a battery that increases the electric potential energy of 0.060 CC of charge by 0.70 JJ as it moves it from t

he negative to the positive terminal?
Physics
2 answers:
Evgen [1.6K]3 years ago
5 0

Answer:

emf = 11.667 V

Explanation:

Given: charge q = 0.060 C, electric potential energy E =0.70 J,

Solution :

by definition 1 volt = 1 joule per coulomb

so Voltage = emf = E/C

emf =  0.70 J / 0.060 C

emf = 11.667 V

Anastasy [175]3 years ago
5 0

Answer:

11.67 V.

Explanation:

Workdone by a charge, W = q × v

Where,

W = 0.7 J

q = 0.06 C

v = W/q

= 0.7/0.06

= 11.67 V

Emf = 11.67 V

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Answer:

the internal resistance of the cell is 0.1 ohm.

Explanation:

Given;

p.d at the terminals of a battery at no load, E₁ = 25 V

p.d at the terminals of a battery at a load, E₂ = 24 V

current through the circuit, I = 10 A

The potential drop across the circuit, V = E₁ - E₂

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                                                               = 1 V

The internal resistance of the cell is calculated as follows;

r = V/I

r = 1 / 10

r = 0.1 ohm

Therefore, the internal resistance of the cell is 0.1 ohm.

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