A battery delivering a current of 55.0 A to a circuit has a terminal voltage of 22.4 V. The electric power being dissipated by the internal resistance of the battery is 37 W. Find the emf of the battery.

Respuesta :

The emf of the battery is 23.071 V.

What is emf?

The emf (electromotive force) is defined as the original potential applied across a circuit when there is no internal resistance.

The emf is given by the formula,

E=V+I*r

where E is the emf of the battery, V is the potential applied across the circuit, I is the current through the circuit and r is the internal resistance of the battery.

What is power?

The power is the product of the square of the current and the resistance. The formula of the power P for the internal resistance r is,

P=I^(2)*r

Given I=55.0 A and P=37 W, substitute I=55.0 A and P=37 W in the above formula and solve it to calculate r.

37 = (55.0)^2*r

r=37/ (55.0)^2

r=0.0122 ohm

Given V=22.4 V, substitute  I=55.0 A, r=0.0122 ohm, and V=22.4 V in the formula of emf to calculate the value of emf.

E=22.4+(55.0)*(0.0122)

E=23.071 V

Learn more about emf here:

https://brainly.com/question/14821588

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