Remember that internal resistance lowers the EMF of the battery. We need to calculate the voltage drop that would occur inside the battery, and then subtract that from the “stated” voltage to determine the terminal potential of the battery.

We first need the current supplied by the battery. We can use the formula V = IR because we have the voltage drop across the circuit (9V) and can calculate the equivalent resistance.

By taking the inverse of the equation, we can see that R_A4 is equal to 2Ω.

R_eq = R_A4 + R₁ = 2Ω + 2Ω = 4Ω

Now, using V=IR, we can find the current in the circuit.

V = IR

I = V/R = 9V/4Ω = 2.25A

Plugging in this value with the internal resistance of the battery gives us the voltage decrease.

V = IR = (2.25A)(0.9Ω) = 2.025V

Terminal Potential = 9V – 2.025V = 6.975V

As we can see, battery companies are interested in keeping the internal resistance of batteries at a minimum, because it lowers the overall terminal potential of the battery.

For this question we need to know that the voltage drop is equal through all paths of a parallel circuit; therefore we can realize that the voltage drop through both resistors are independent of each other. The current through resistor A will decrease with the removal of resistor B. Electrons want to travel through the path of least resistance.

Given that resistor R = 2 Ω, we can see that the parallel aspect of the circuit has a resistance of 2 Ω as a whole, which means the total effective resistance of the circuit is 4 Ω. The potential difference drops 2V over the first resistor, then 2 V through the parallel aspect of the circuit. Potential difference, or voltage, drops through an equal amount in all branches of resistors in parallel. Therefore the remaining 2 V of potential difference is dropped by electrons moving through both branches of the parallel resistor, and the voltage drop through the 4 Ω resistor is 2V.

Given that

V(q) = U

When V is voltage, U is electrical potential energy and q is charge, we can solve by plugging in 4 for V and -2 for q. Also, we must understand that the electric potential energy of a particle decreases as it moves from an area of higher energy to one of lower energy. For an electron, it has higher energy in the negative terminal of the battery than it does in when getting to the positive terminal.

This question asks us directly about one of Kirchhoff’s Loop Laws—more specifically the second law that states that the sum of the voltage around any loop must be equal to zero.

ΣV_circuit = 0V

Because the battery provides 9V, the voltage drop across the circuit must be 9V.

As an aside, the other Kirchhoff Law, the first law, states that current into a junction must equal the current that exits the junction, due to conservation of charge.

First, we need to determine the voltage drop across R₁, and then we can subtract that from the 9V battery to determine the voltage drop across R_A. Given that R_A and R₄ are in parallel, we know that the voltage drop across each is the same.

The voltage drop across R₁ can be calculated using the I_circuit (all the current generated by the battery’s potential difference must pass through R₁ because it is in direct series with the battery), and the resistance of the resistor.

I can be calculated by V = IR because we have the voltage drop across the circuit (9V) and can calculate the equivalent resistance.

By taking the inverse of the equation, we can see that R_A4 is equal to 2Ω.

R_eq = R_A4 + R₁ = 2Ω + 2Ω = 4Ω

Now, using V=IR allows us to find the current in the circuit.

V = IR

I = V/R = 9V/4Ω = 2.25 A

Plugging this value in, we can find the voltage drop across R₁.

V_R1 = IR = (2.25 A)(2Ω) = 4.5 V

Now we can determine the voltage drop across the parallel resistors by subtracting the voltage drop across R₁ from the battery.

V_A = 9V – 4.5V = 4.5V

Because R_A and R₄ are in parallel, the voltage drop across R₄ is also 4.5V

Ohm's Law states that . We are given the current and resistance, allowing us to solve for the voltage.

We also know that . Now that we have the voltage, we can solve for the power.

Finally, we know that . Using this, we can solve for the energy (or work) that is generated by the circuit.

Recall that circuit elements (such as resistors) connected in parallel have the same potential difference; therefore, resistor A and resistor B must have the same potential difference. Since we have information about the potential difference and the resistance of the two resistors, it would be best to use the equation for power in terms of voltage and resistance.

Since the potential difference is constant for both resistors, the power will only depend on the resistance. The equation implies that the power decreases when resistance is high; therefore, resistor A will have lower power dissipation than resistor B because resistor A has the greater resistance.

Resistors connected in series will have the same current flowing through them, but will have different potential differences. Since we are concerned with the differences in potential difference and resistance, we need to use the power dissipation equation in terms of voltage and resistance:

The equation implies that there will be a greater power dissipation when the potential difference increases and the resistance decreases. A resistor undergoing greater power dissipation will most likely have a higher potential difference and lower resistance.

Power dissipation is defined as the ability of a circuit element (such as a resistor) to convert the electrical energy to other forms of energy, such as heat or mechanical energy. This means that a decrease in power dissipation will convert less of the electrical energy to heat and mechanical energy. Statement I is false (and therefore a correct answer choice).

Electrical power can be written in terms of voltage () and resistance (), voltage and current (), or resistance and current. Since we are only concerned with the effects of resistance on power dissipation, we will use the equation for power in terms of voltage and resistance, and resistance and current:

Notice that resistance is in the numerator in the first equation and in the denominator in the second equation; therefore, the power doesn’t always decrease when resistance increases. It depends on the equation used to calculate the power. Statement II is false (and therefore a correct answer choice).

Amperes is the unit of current and volts is the unit of voltage; therefore, we need to find the equation for power in terms of voltage and current to determine if power can be measured in volt-amperes. The equation for power in terms of voltage and current is:

Power can be measured in the units of volt-amperes (VA). Statement III is true (and therefore an incorrect answer option).