A battery delivering a current of 55.0 A to a circuit has a terminal voltage of 21.8 V. The electric power being dissipated by the internal resistance of the battery is 36 W. Find the emf of the battery.
Vi * I = 36 W.
Vi * 55 = 36
Vi = 0.655 V0lts. = Internal voltage drop.
E = 21.8 + 0.655 = 22.5 V. = emf of battery.
To find the electromotive force (emf) of the battery, we can use the equation:
emf = terminal voltage + internal resistance * current
Given that the terminal voltage of the battery is 21.8 V and the current is 55.0 A, we can substitute these values into the equation:
emf = 21.8 V + internal resistance * 55.0 A
We’re also given that the power dissipated by the internal resistance is 36 W. The power is given by the equation:
power = (internal resistance) * (current)^2
Substituting the given values:
36 W = (internal resistance) * (55.0 A)^2
Simplifying:
36 W = (internal resistance) * 3025 A^2
We can solve for the internal resistance:
internal resistance = 36 W / 3025 A^2
Now that we have the internal resistance, let's substitute this value back into the equation for emf:
emf = 21.8 V + (internal resistance) * 55.0 A
emf = 21.8 V + (36 W / 3025 A^2) * 55.0 A
Simplifying:
emf = 21.8 V + (36 W / 55.0) A
emf = 21.8 V + 0.654545... V
emf ≈ 22.45 V
Therefore, the emf of the battery is approximately 22.45 V.