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

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

Step-by-step 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=(V)/(R)` (2)

using (2) on (1):

`V=\xi-(V*r)/(R)`

solving for V:

`V+(V*r)/(R)=\xi`

`V=(\xi)/(1+(r)/(R))` (3)

R=100 Ohm

`V=(12.2)/(1+(1.9)/(100))=11.97 V`

R=10 Ohm

`V=(12.2)/(1+(1.9)/(10))=10.25 V`

R=2 Ohm

`V=(12.2)/(1+(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=(V)/(R)=(11.97)/(100)=0.12 A`

R=10 Ohm

`I=(V)/(R)=(11.97)/(10)=1.20 A`