MCAT Physical - Resistors and Resistance

What is the resistance through resistor R given that the current through the circuit is 1 A?

2 Ω

1 Ω

0.5 Ω

4 Ω

Explanation

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

What is the internal resistance of a battery that has a potential of when a current runs through it?

$5\Omega$

$15\Omega$

$3\Omega$

$30\Omega$

Explanation

The equation for a battery with internal resistance is $V = {emf} - IR_{internal}$

Using this equation, we can solve for the resistance.

$R = \frac{{emf}-V}{I} = \frac{25V-10V}{3A} = 5\Omega$

Which of these statements is false?

In the case of parallel resistors, the greatest current flow will occur across the highest resistor.
In the case of parallel resistors, the greatest current flow will occur across the lowest resistor.
In the case of series resistors, the current flow is the same through each resistor.
In the case of series resistors, the voltage drop is greatest across the highest resistor.
All are false.

Explanation

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.

During the cold winter months, some gloves have the ability to provide extra warmth due to an internal heating source. A simplified circuit, similar to those in electric gloves, is comprised of a 9V battery with no internal resistance and three resistors as shown in the image below.

Screen_shot_2013-10-09_at_11.08.15_pm

What is the net resistance of the circuit?

$4\Omega$

$2.5\Omega$

$2\Omega$

$9\Omega$

$11\Omega$

Explanation

First we need to combine the resistors in parallel.

$\frac{1}{R_{A4}}=\frac{1}{R_A}+\frac{1}{R_4}$

$\frac{1}{R_{A4}}=\frac{1}{6\Omega }+\frac{1}{3\Omega}=\frac{1}{2}\Omega$

By taking the inverse of the equation, we can see that RA4 is equal to 2Ω.

Now that we have simplified the two resistors in parallel, we need to determine how the Req of the two parallel resistors and the resistance due to R1 are arranged. We can see that these resistors are now in series, thus we can directly add their resistances together to get the overall resistance of the circuit.

Req = RA4 + R1 = 2Ω + 2Ω = 4Ω

Remember as a general rule, the equivalent resistance of resistors in series is an arithmetic sum of their individual resistances.

During the cold winter months, some gloves have the ability to provide extra warmth due to an internal heating source. A simplified circuit, similar to those in electric gloves, is comprised of a 9 V battery with no internal resistance and three resistors as shown in the image below.

Screen_shot_2013-10-09_at_11.08.15_pm

All resistors in series share the same __________, while all resistors in parallel share the same __________.

current . . . voltage

voltage . . . current

current . . . potential energy

potential energy . . . voltage

Explanation

As a common rule to all circuits, all resistors in series share the same current (due to conservation of charge) and all resistors in parallel share the same voltage. This information is very helpful in determining the voltage drops across sets of resistors and the current that flows through them.

Use the following information to answer questions 1-6:

The circulatory system of humans is a closed system consisting of a pump that moves blood throughout the body through arteries, capillaries, and veins. The capillaries are small and thin, allowing blood to easily perfuse the organ systems. Being a closed system, we can model the human circulatory system like an electrical circuit, making modifications for the use of a fluid rather than electrons. The heart acts as the primary force for movement of the fluid, the fluid moves through arteries and veins, and resistance to blood flow occurs depending on perfusion rates.

To model the behavior of fluids in the circulatory system, we can modify Ohm’s law of V = IR to ∆P = FR where ∆P is the change in pressure (mmHg), F is the rate of flow (ml/min), and R is resistance to flow (mm Hg/ml/min). Resistance to fluid flow in a tube is described by Poiseuille’s law: R = 8hl/πr4 where l is the length of the tube, h is the viscosity of the fluid, and r is the radius of the tube. Viscosity of blood is higher than water due to the presence of blood cells such as erythrocytes, leukocytes, and thrombocytes.

The above equations hold true for smooth, laminar flow. Deviations occur, however, when turbulent flow is present. Turbulent flow can be described as nonlinear or tumultuous, with whirling, glugging or otherwise unpredictable flow rates. Turbulence can occur when the anatomy of the tube deviates, for example during sharp bends or compressions. We can also get turbulent flow when the velocity exceeds critical velocity vc, defined below.

vc = NRh/ρD

NR is Reynold’s constant, h is the viscosity of the fluid, ρ is the density of the fluid, and D is the diameter of the tube. The density of blood is measured to be 1060 kg/m3.

Another key feature of the circulatory system is that it is set up such that the organ systems act in parallel rather than in series. This allows the body to modify how much blood is flowing to each organ system, which would not be possible under a serial construction. This setup is represented in Figure 1.

Circulatory_system_circuit

Due to the parallel circuit, the circulatory system has the ability to modify the flow of blood to different organs in the body. The body does this by dilating or constricting vessels to change resistance, and thus the flow of blood.

Assume that, in the diagram in Figure 1, all of the resistances are initially equal. In order to direct more blood flow through R3, the body reduces the resistance to R3 by dilating vessels, and increases R1 and R2 equally by restricting vessels. Assume Rb is the resistance to blood flow through the body from the left ventricle, and it remains constant. What is the value of R3 in terms of Rb?

R3 = 2Rb

R3 = Rb

R3 = 4Rb

R3 = 8Rb

Explanation

We need to use the parallel resistance equation.

$\frac{1}{R_b}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}$

R4 is not included since that is part of a different circulatory circuit; we are only concerned with blood flow to the body, not the lungs. We are told that they all start with the same resistance, so R1 = R2 = R3, which we can just call R.

Now, given V = IR, in order for flow to double across a parallel resistor, since voltage is constant, resistance must decrease by a factor or two. Now R3 = (1/2)R.

We can rewrite the initial equation and solve for R.

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

We can now solve for R3 using the relationship between R3 and R discussed earlier.

$R3=\frac{1}{2}R=2R_b$

If each resistor in the diagram above has resistance , determine the equivalent resistance of the circuit.

$\frac{41}{15}R$

$\frac{15}{41}R$

$\frac{1}{9}R$

Explanation

Resistors add differently, depending on their orientation in series or in parallel. The three resistors furthest to the right are in series and can be simplified to $\small R+R+R = 3R$ .

Now this "" resistor is in parallel with the resistor adjacent to it. For parallel circuit elements $\small \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$ . Combining the four rightmost resistors gives the below calculation.

$\frac{1}{R_{1234}}=\frac{1}{R_{123}}+\frac{1}{R_4}=\frac{1}{3R}+\frac{1}{R}$

$\frac{1}{R_{1234}}=\frac{4}{3R}\rightarrow R_{1234}=\frac{3}{4}R$

This method can be repeated for the next sets of resistors in series and in parallel.

For the final calculation, $\frac{11}{15}R + R + R = \frac{41}{15}R$ .

A battery is connected to a circuit which contains two $5\Omega$ resistors connected in parallel. What is the current throughout the circuit?

Explanation

We know the total voltage in the circuit is . We need to determine the total resistance within the circuit. Since we know there are two resistors connected in parallel we use the following formula:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

Two resistors with resistances of $5\Omega$ each:

$\frac{1}{R_{eq}}=\frac{1}{5\Omega}+\frac{1}{5\Omega} = \frac{2}{5}\Omega$

$R_{eq}=\frac{5}{2}=2.5\Omega$

Now we use another formula to determine the current using voltage and resistance:

Rearranged to:

$I=\frac{V}{R}= \frac{25V}{2.5\Omega }=10A$

During the cold winter months, some gloves have the ability to provide extra warmth due to an internal heating source. A simplified circuit, similar to those in electric gloves, is comprised of a 9 V battery with no internal resistance and three resistors as shown in the image below.

Screen_shot_2013-10-09_at_11.08.15_pm

What happens to the energy that is dissipated by the resistors?

Dissipated as heat

Dissipated as light

Not enough information provided to determine

Stored as potential energy

Explanation

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.

During the cold winter months, some gloves have the ability to provide extra warmth due to an internal heating source. A simplified circuit, similar to those in electric gloves, is comprised of a 9 V battery with no internal resistance and three resistors as shown in the image below.

Screen_shot_2013-10-09_at_11.08.15_pm

What is the total resistance of the two resistors in parallel (RA and R4)?

2Ω

9Ω

1/2Ω

9/2Ω

Explanation

First, we need to determine how the resistors are being added. As we can see in the circuit diagram, the two resistors are being added in parallel. In parallel, we should expect an equivalent resistance lower that that of each individual resistor.

$\frac{1}{R_{eq}}=\frac{1}{R_A}+\frac{1}{R_4}$

$\frac{1}{R_{eq}}=\frac{1}{6\Omega }+\frac{1}{3\Omega}=\frac{1}{2}\Omega$

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

Remember as a general rule, the equivalent resistance of resistors in parallel is lower than that of either resistor in the parallel circuit.

Resistors and Resistance

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