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).

The question is asking for the amount of energy lost per unit of time; therefore, we need to solve for the power dissipation. We are given the resistance and the voltage. This means that we use the power dissipation equation in terms of resistance and voltage:

Solving for power gives us:

Recall that watts is the SI unit for power, but it can be also written as Joules per second. Since we are solving for Joules lost per hour, we need to convert our units:

This means that 108J of electrical energy is lost every hour due to power dissipation in the resistor.  

In order to solve this problem, we must understand that the resistance of parallel resistors is added inversely and that series resistors are added directly.

Series

Parallel

Ohm’s Law

We can first find the total resistance of the circuit to be 4 Ω using ohm's law . Then by subtracting the resistor in series (2 Ω) from the total resistance, we can set the remaining parallel resistors equal to 2 Ω. Finally, we can use the parallel resistor equation to find R. Remember that you can simply count the two resistors in series on the far right branch of the circuit (R and 2 Ω) as one number in the parallel resistor formula giving us

we can solve for this to find that R = 2 Ω

Choice 1 is a false statement, and therefore the correct response. Electrical potential energy (voltage), like water, always flows downhill along the path of least resistance. In the situation of parallel resistors, the greatest current flow will be through the lowest resistor.

(Mnemonic: Think of an ultra-low resistor = a wire. The current would preferentially flow through the wire rather than, for example, a one million Ohm resistor.)  In an electrical circuit, the current entering the circuit from one pole of the battery and the current re-entering the other pole of the battery must be the same, although the voltage is exhausted by passing through the various resistors in the circuit in direct proportion to the degree of resistance each causes.

Resistors serve to drop voltage over a certain distance by providing friction for electrons to move against. As with other types of friction, including kinetic and static contact friction and air resistance, energy is lost as heat to the surrounding environment. The same is true in the circuits of these winter gloves. The resistors allow the energy provided by the battery to be dissipated as heat, warming the person's hands during the cold winter months.