The terminal voltage of a battery is 5.0V when 0.55A current is drawn from it, whereas the terminal voltage is 5.7V when a current of 0.180A is drawn. The emf of the battery is?
The formula I used was
V_(ab) = Emf - I*r
First finding the r by taking the differences of voltage and the currents, which then for r I got is .7V/.37A = 1.89 Ohms. and then inserting it back to the formula above I get the Emf to be 1.39. Which I am not sure if it is right. Any help will be appreciated.
emf - (internal resistance * current) = terminal voltage
(Ri * .55) + 5.0 = (Ri * .18) + 5.7
emf - (1.9 * .55) = 5.0
Ah I see, thank you so much.
To find the electromotive force (emf) of the battery, you can use the formula V(ab) = Emf - I*R, where V(ab) is the terminal voltage, I is the current drawn from the battery, and R is the internal resistance of the battery.
Given the following data:
- V(ab) = 5.0V when I = 0.55A
- V(ab) = 5.7V when I = 0.180A
Let's calculate the internal resistance (R) first. Using the formula R = (V1 - V2) / (I1 - I2), where V1 and V2 are the voltage readings and I1 and I2 are the corresponding current readings, we get:
R = (5.0V - 5.7V) / (0.55A - 0.180A)
= -0.7V / 0.37A
≈ -1.89 Ω
The negative sign indicates that the internal resistance is in the opposite direction of the current flow. However, resistances cannot be negative, so let's take the absolute value and consider it as a positive value.
Now, we can use the formula V(ab) = Emf - I*R to calculate the emf (Emf) of the battery. Plugging in the values we have:
5.0V = Emf - (0.55A * 1.89Ω)
5.0V + (0.55A * 1.89Ω) = Emf
Emf ≈ 5.0V + 1.04V
Emf ≈ 6.04V
So, the estimated emf of the battery is approximately 6.04 volts (V).