Circuits - MCAT Chemical and Physical Foundations of Biological Systems

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.

The power consumed by a capacitor is given by

P=\frac{1}{2}$qV

After finding power, we can calculate the energy used as we are told that the capacitor took 5 seconds to charge. The units of power are watts which can be further broken down into joules per second. By multiplying the time taken to charge the capacitor by the power, we can find the total energy required to charge the capacitor.

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.

ΣVcircuit = 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 RA. Given that RA 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 Icircuit (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 RA4 is equal to 2Ω.

Req = RA4 + 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.

VR1 = 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.

VA = 9V – 4.5V = 4.5V

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

This question asks us about the power of the circuit, meaning the amount of energy per unit time. The power equation is P = IV. V = IR can be substituted in to allow P to be calculated from a number of parameters.

P = IV = I2R = V2/R

To solve for P, 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 RA4 is equal to 2Ω.

Req = RA4 + R1 = 2Ω + 2Ω = 4Ω

Now, using V=IR, we can solve for the current.

V = IR

I = V/R = 9V/4Ω = 2.25A

Now, we can use the current we calculated and the voltage that is dissipated across the circuit to calculate the power.

P = IV = (2.25 A)(9V) = 20.25W

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 RA4 is equal to 2Ω.

Req = RA4 + 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.

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.

The brightness of each bulb depends on its power, or how much energy it dissipates per unit time. This can be calculated using the equation:

If the bulbs are in parallel with each other, the bulb with lower resistance will receive more current, since current will tend toward the path of least resistance. Current is squared in the power equation, so it is of greater importance in determining the power. The bulb with greater current and lower resistance will thus have more power and will shine more brightly.

The formula we can use here is the power formula that involved both resistance and voltage:

We are given the voltage and power, allowing us to solve for the resistance.

Electric power is expressed in a number of ways. We will need to use the relationship that involves current and resistance:

We are given the current and resistance, allowing us to calculate the power.

Alternative current (AC) voltage is governed by the RMS (root-mean square) voltage law. This means that the RMS voltage of the circuit will be , but the actual voltage will fluctuate between two values. These values are determined by the equation:

Using the values from our question, we can find the maximum voltage value.

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.

To find resistance from voltage and current, we will need to use Ohm's law:

We are given the voltage and current, allowing us to solve for the resistance.

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
$R_{total}$= $R_{1}$ + $R_{2}$ + $R_{3}$

Parallel
$$\frac{1}{R_{total}$$}= $$\frac{1}{R_{1}$$} + $$\frac{1}{R_{2}$$} + $$\frac{1}{R_{3}$$}

Ohm’s Law
V=iR

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

$\frac{1}{2}$= $\frac{1}{4}$ + $\frac{1}{2 + R}$

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.