Question

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.

Answer 1

Answer:

R=100 Ohm, V=11.97 volts and I=0.12 amperes

R=10 Ohm, V=10.25 volts and I=1.20 amperes

R=2 Ohm, V=6.26 volts

Explanation:

The potential difference (voltage) of a battery with internal resistance is:

$V=\xi-Ir$ (1)

with $\xi$ 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:

$I=\frac{V}{R}$ (2)

using (2) on (1):

$V=\xi-\frac{V*r}{R}$

solving for V:

$V+\frac{V*r}{R}=\xi$

$V=\frac{\xi}{1+\frac{r}{R}}$ (3)

R=100 Ohm

$V=\frac{12.2}{1+\frac{1.9}{100}}=11.97 V$

R=10 Ohm

$V=\frac{12.2}{1+\frac{1.9}{10}}=10.25 V$

R=2 Ohm

$V=\frac{12.2}{1+\frac{1.9}{2}}=6.26 V$

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

$I=\frac{V}{R}=\frac{11.97}{100}=0.12 A$

R=10 Ohm

$I=\frac{V}{R}=\frac{11.97}{10}=1.20 A$