Question

PLEASE HELP ME ASAP

Answer 1

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:

$V=I\,\,R\\6\,V=I\,(5\,\Omega)\\I=\frac{6}{5} \,amps\\I=1.2\,\,amps$

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:

$\frac{i}{R_e} =\frac{1}{6\,\Omega} + \frac{1}{6\,\Omega} =\frac{1}{3\,\Omega} \\R_e=3\,\Omega$

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:

$V=I\,\,R\\1.5\,V=I\,(5\,\Omega)\\I=\frac{1.5}{5} \,amps\\I=0.3\,\,amps$

c. The potential difference across the battery terminals must be the battery's EMF minus the potential drop in its internal resistance:

$1.5\,V - (0.3\,\,amps)\,(2\,\Omega)=(1.5-0.6)\,V=0.9\,V$