Answer:
1. Current in the circuit; 1.2 Amps
See attached image for the circuit.
2. Equivalent resistor = 3 Ω
I = 0.3 amps
Potential difference across the battery terminals is: 0.9 V
Explanation:
Part 1.
The internal resistance of 2 ohms is simply added to the circuit in series as shown in the attached image.
Since now we have two resistances in series (2 ohms and 3 ohms) the total of this series combination is 5 ohms. Using Ohm's law, we can derive the current running through the circuit:
[tex]V=I\,\,R\\6\,V=I\,(5\,\Omega)\\I=\frac{6}{5} \,amps\\I=1.2\,\,amps[/tex]
Part 2.
Now we have a 1.5 V battery with a 2 ohm internal resistance, connected to two identical 6 ohm resistors.
a. The equivalent resistance presented by the two resistors in parallel:
[tex]\frac{i}{R_e} =\frac{1}{6\,\Omega} + \frac{1}{6\,\Omega} =\frac{1}{3\,\Omega} \\R_e=3\,\Omega[/tex]
b. Now the circuit can be represented by a 2 ohm resistor (internal battery resistance) plus a 3 ohm parallel equivalent resistor in series. That is a 5 ohm total resistance. Then Ohm's law becomes:
[tex]V=I\,\,R\\1.5\,V=I\,(5\,\Omega)\\I=\frac{1.5}{5} \,amps\\I=0.3\,\,amps[/tex]
c. The potential difference across the battery terminals must be the battery's EMF minus the potential drop in its internal resistance:
[tex]1.5\,V - (0.3\,\,amps)\,(2\,\Omega)=(1.5-0.6)\,V=0.9\,V[/tex]
