A non-ideal 12.2 V battery is connected across a resistor R. The internal resistance of the battery is 1.9Ohm. Calculate the potential difference across the resistor for the following values. R = 100 Ohm Express the potential difference in volts to two significant figures. R = 10 Ohm Express the potential difference in volts to two significant figures. R = 2 Ohm Express the potential difference in volts to two significant figures. Find the current through me battery for R = 100 Ohm Express your answer in amperes to two significant figures. Find the current through me battery for R = 10 Ohm Express your answer in amperes to two significant figures.

with [tex]\xi [/tex] the electromotive force (the voltage the batteries say to has) , I the current and r the internal resistance. By Ohm's law the current that passes through the resistor is:

[tex]I=\frac{V}{R} [/tex] (2)

using (2) on (1):

[tex]V=\xi-\frac{V*r}{R}[/tex]

solving for V:

[tex] V+\frac{V*r}{R}=\xi[/tex]

[tex]V=\frac{\xi}{1+\frac{r}{R}} [/tex] (3)

R=100 Ohm

[tex] V=\frac{12.2}{1+\frac{1.9}{100}}=11.97 V[/tex]

R=10 Ohm

[tex] V=\frac{12.2}{1+\frac{1.9}{10}}=10.25 V[/tex]

R=2 Ohm

[tex] V=\frac{12.2}{1+\frac{1.9}{2}}=6.26 V[/tex]

Because we have now the values of I on the circuit (is the same through all the components because is a series circuit)

We use back substitution on (1) to find the current:

R=100 Ohm

[tex]I=\frac{V}{R}=\frac{11.97}{100}=0.12 A [/tex]

R=10 Ohm

[tex]I=\frac{V}{R}=\frac{11.97}{10}=1.20 A [/tex]