Calculating Terminal Voltage, Power Dissipation, Current, and Resistance: Terminal Voltage and Load

The analysis above gave an expression for current when internal resistance is taken into account. Once the current is found, the terminal voltage can be calculated using the equation $V=\text{emf}-\text{Ir}$. Once current is found, the power dissipated by a resistor can also be found.

Entering the given values for the emf, load resistance, and internal resistance into the expression above yields

Enter the known values into the equation $V=\text{emf}-\text{Ir}$ to get the terminal voltage:

The terminal voltage here is only slightly lower than the emf, implying that $\text{10}\text{.}0\Omega$ is a light load for this particular battery.

Similarly, with $R_{\text{load}}=0\text{.}\text{500}\mathrm{<mspace\; width="0.15em"><mo\; stretchy="false">\Omega </mo></mspace>}$, the current is

The terminal voltage is now

This terminal voltage exhibits a more significant reduction compared with emf, implying $0\text{.}\text{500}\Omega$ is a heavy load for this battery.

The power dissipated by the $0\text{.}\text{500 - Ω}$ load can be found using the formula $P=I^{2}R$. Entering the known values gives

Note that this power can also be obtained using the expressions $\frac{V^2}{R}$ or $\text{IV}$, where $V$ is the terminal voltage (10.0 V in this case).

Here the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load resistance. As before, we first find the current by entering the known values into the expression, yielding

Now the terminal voltage is

and the power dissipated by the load is

We see that the increased internal resistance has significantly decreased terminal voltage, current, and power delivered to a load.