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MCAT > MCAT Physics > Circuits > Flashcards

Circuits Flashcards

From the basics of voltage and current to in-depth circuit calculations, use these cards to master the topic of circuits as it appears on the MCAT.

1
Q

Define:

electric current (I)

A

It is the directional flow of charge through a conducting medium.

Though the moving charges in a circuit are typically electrons, the conventional direction of current always follows positive charge flow.

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2
Q

What formula and SI unit are associated with current?

A

SI unit is ampere (A), it is associated by:

I = Q / t

Where:

Q = charge transferred (coulombs)
t = time (s)

Ampere represents 1 C/s.

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3
Q

If electrons are moving along a wire from left to right, in what direction is current flowing?

A

Conventional current is moving from right to left.

By scientific convention and on the MCAT, current always refers to the flow of positive charge. It will always be opposite to the flow of electrons.

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4
Q

By what proportion will the current change if the amount of charge transferred doubles and the time is halved?

A

The current will increase by a factor of 4.

I0 = Q / t
Under the new conditions, Q’ = 2Q and t’ = (1/2)t.
I’ = Q’ / t’
= 2Q / (1/2)t = 4(Q / t)
= 4I0

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5
Q

Define:

battery

A

An electrical battery (or electrochemical cell) is a device that produces a flow of electrons from anode to cathode.

The necessary electrical energy is created by redox chemical reactions that are undergone within the cell.

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6
Q

Define:

electromotive force

(EMF)

A

It is the ability to transfer a unit of charge over the electrical potential difference between two electrodes.

Note that EMF is measured in volts and is not an actual force.

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7
Q

Define:

electric potential energy

A

Electric potential energy (or electrostatic potential energy) is the energy, either for repulsion or attraction, between specific charges.

Technically, electric potential can be calculated between electric fields as well, but this is beyond the scope of the MCAT.

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8
Q

What formula and SI unit are associated with electric potential energy?

A

SI unit is joule (J) it is associated by:

U = kQq/r

Where:

k = Coulomb’s constant (9x109 Nm2/C2)
Q = source charge (coulombs)
q = test charge (coulombs)
r = radius between charges (m)

As with all forms of energy, it is defined as 1 N*m.

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9
Q

What does the sign of the electric potential energy indicate about the charges and force between them?

A
  • A positive energy exists when both charges have the same sign (both positive or both negative). This will cause a repulsive force between the two charges.
  • A negative energy exists when the charges have opposite signs. This will cause an attractive force between the two charges.
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10
Q

Define:

voltage

A

Voltage, or electric potential, is the energy per unit charge necessary to move a test charge within the electric field of a source charge.

In circuits, voltage is often used interchangeably with EMF, or electromotive force.

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11
Q

What formula and SI unit are associated with voltage?

A

SI unit is volts (V), it is associated by:

V = kQ/r

Where:

k = Coulomb’s constant (9x109 Nm2/C2)
Q = primary charge (coulombs)
r = distance between charges (m)

Voltage is measured in 1 V is equal to 1 J/C.

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12
Q

A test charge begins at a distance r from a central charge Q. By what proportion will the voltage change if the charge is now moved twice as far away?

A

The voltage will decrease by 1/2.

V0 = kQ/r
Under the new conditions, r’ = 2r, while k and Q are constant.

V’ = kQ/r’ = kQ/2r
(kQ/r)(1/2) = V0 / 2

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13
Q

Define:

resistance

A

It is a measure of how difficult it is to pass current through an object. A number of factors affect resistance, including cross-sectional area, length, and resistivity of the material.

Though all materials have some resistance at room temperature, supercooled superconductors have a resistance of zero.

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14
Q

What quantity is generally measured in ohms?

A

Resistance is meaured in ohms (Ω), its SI unit.

1 Ω represents 1 J*s / C2.

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15
Q

What formula can be used to calculate the total resistance of an object?

A

R = ρL / A

Where:

ρ = resistivity of the material (Ωm)
L = length of the object (m)
A = cross-sectional area of the object (m2)

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16
Q

By what factor will total resistance change when the length of a resistor is doubled?

A

The total resistance will also double.

R0 = ρL / A
Under the new conditions, L’ = 2L.

R’ = ρL’ / A
ρ(2L) / A = 2(ρL / A) = 2R0

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17
Q

By what factor will total resistance change when the cross-sectional area of a resistor is doubled?

A

The total resistance will decrease by 1/2.

R0 = ρL / A
Under the new conditions, A’ = 2A.

R’ = ρL / A’
ρL / 2A = (1/2)(ρL / A) = R0 / 2

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18
Q

By what factor will total resistance change if both the length and the cross-sectional area of a resistor double?

A

The total resistance will remain constant.

R0 = ρL / A
Under the new conditions, L’ = 2L and A’ = 2A.

R’ = ρL’ / A’
ρ(2L)/(2A) = ρL / A = R0

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19
Q

By what factor will the total resistance of a wire change if both its length and radius are doubled?

A

The total resistance will decrease by 1/2.

R0 = ρL / A
Under the new conditions, L’ = 2L. However, A’ = 4A, not 2A, since area is proportional to the square of the radius.

R’ = ρL’ / A’
ρ(2L) / 4A = ρL / 2A = R0 / 2

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20
Q

A real battery will generally produce a voltage that is slightly less than its predicted ideal voltage. What property explains this effect?

A

Like any element of a circuit, any battery will contain its own internal resistance. This resistance provides the difference between the battery’s theoretical voltage and the actual value that it provides.

Occasionally, internal resistance may be referred to as impedance. Simply remember that factors like battery size, chemical makeup, and current load can impede the ideal voltage.

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21
Q

What is resistivity, and in which units is it measured?

A
  • Also known as ρ, is the measurement of a certain material’s resistance to the flow of current.
  • It is measured in Ω*m.

Different materials have different resistivity values.

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22
Q

What is the formula for Ohm’s law?

A

V = IR.

Where:

V = voltage (V)
I = current (A)
R = resistance (Ω)

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23
Q

In a circuit with constant current flow, what happens to the resistance when the voltage is doubled?

A

The resistance also doubles.

According to Ohm’s law, V = IR. Since current is held constant, voltage and resistance are directly proportional.

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24
Q

In a fixed-resistance circuit, the current suddenly decreases to half of its original value. What change must have been made to the voltage?

A

The voltage must have also dropped by one-half.

According to Ohm’s law, V=IR. Since resistance is held constant, voltage and current are directly proportional.

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25
Q

In a fixed-voltage circuit, the current suddenly doubles. What change must have been made to the resistance?

A

The resistance must have dropped by one-half.

According to Ohm’s law, V = IR. Since voltage is held constant, resistance and current are inversely proportional.

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26
Q

What does this symbol stand for in a circuit diagram?

A

battery

Batteries will always have a larger positive terminal and a smaller negative terminal.

Conventional current always flows around the circuit from the positive to the negative terminal.

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27
Q

What does this symbol stand for in a circuit diagram?

A

resistor

It provides resistance, which impedes current flow and is measured in ohms.

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28
Q

What does this symbol stand for in a circuit diagram?

A

capacitor

It provides capacitance, which stores charge and is measured in farads.

29
Q

What does this symbol stand for in a circuit diagram?

A

switch

The switch shown here is open.

Switches can be open to prevent current flow around the circuit or closed to allow current flow.

30
Q

What does this symbol stand for in a circuit diagram?

A

ground

The ground is the physical end of the circuit path and represents zero current or zero voltage.

If no ground symbol is given, the negative terminal of the battery is assumed to be the ground.

31
Q

Describe the difference between series and parallel resistors, with respect to their appearance in a circuit diagram.

A
  • Resistors in parallel are located on diverging branches of the circuit, as shown in the image on the left. A single charge will pass through either one or the other to complete the circuit.
  • Resistors in series follow each other on a continuous pathway, as shown in the image on the right. A single charge must travel consecutively through both to complete the circuit.
32
Q

For resistors in series, what formula can be used to calculate the total resistance of a circuit?

A

Rtotal = R1 + R2 + …

Where:

Rtotal = total resistance (Ω)
R1 = resistance of first resistor (Ω)
R2 = resistance of second resistor (Ω)
… = resistance of additional resistors, similar to above

33
Q

What is the total resistance in the circuit below?

A

Rtotal = 10 Ω

Rtotal = R1 + R2
4 Ω + 6 Ω = 10 Ω

34
Q

For resistors in parallel, what formula can be used to calculate the total resistance of a circuit?

A

1/Rtotal = 1/R1 + 1/R2 + …

Where:

Rtotal = total resistance (Ω), also called equivalent or effective resistance
R1 = resistance of first resistor (Ω)
R2 = resistance of second resistor (Ω)
… = resistance of additional resistors, similar to above

35
Q

What is the total resistance in the circuit below?

A

Rtotal = 5 Ω

1/Rtotal = 1/R1 + 1/R2
1/10 Ω+ 1/10 Ω = 2/10 Ω = 1/5 Ω

36
Q

What is the total resistance in the circuit below?

A

5Ω

First, find the resistance for the resistors in series.
Rseries = 4 Ω + 6 Ω = 10 Ω

Next, find the resistance for the resistors in parallel: the 10 Ω one from the bottom branch and the 10 Ω total for the top branch, calculated above.
1/Rtotal = 1/10 + 1/10 = 2/10 = 1/5
Rtotal = 5 Ω

37
Q

Find the current flowing within the circuit pictured below.

A

1A

First, find the resistance for the resistors in parallel.

1/Rtotal = 1/12 + 1/6 = 3/12 = 1/4
Rtotal = 4 Ω

Then, from Ohm’s law:
I = V/R = 4/4 = 1 A

38
Q

Find the current flowing within the circuit pictured below.

A

2A

First, find the resistance for the resistors in series.
Rseries = 4 Ω + 6 Ω = 10 Ω

Next, find resistance for the resistors in parallel: the 10 Ω one from the bottom branch and the 10 Ω total for the top branch, calculated above.
1/Rtotal = 1/10 + 1/10 = 2/10 = 1/5
Rtotal = 5 Ω

Finally, from Ohm’s law:
I = V/R = 10/5 = 2 A

39
Q

Define:

capacitance

A

It is the measured ability of a material to store charge.

On the MCAT, capacitance is often tested using parallel plate capacitors. For these structures, charge is related to the area of the plates, the distance separating the plates, and the dielectric material between the plates.

40
Q

What quantity is generally measured in farads?

A

capacitance

1 F represents 1 C/V.

Its SI unit is F.

41
Q

What formula can be used to calculate capacitance?

A

C = Q / V

Where:

C = capacitance (farads)
Q = charge stored (coulombs)
V = voltage (V)

42
Q

By what factor will capacitance change if the charge on a capacitor plate doubles while the voltage remains constant?

A

Capacitance will also double.

C0 = Q / V
Under the new conditions, Q’ = 2Q.

C’ = Q’ / V = 2Q / V = 2(Q / V) = 2C0

43
Q

For capacitors in parallel, what formula can be used to calculate the total capacitance of a circuit?

A

Ctotal = C1 + C2 + …

Where:

Ctotal = total capacitance (F)
C1 = capacitance of first capacitor (F)
C2 = capacitance of second capacitor (F)
… = capacitance of additional capacitors, similar to above

44
Q

What is the total capacitance in the circuit below?

A

Ctotal = 10 F

Ctotal = C1 + C2
8 F + 2 F = 10 F

45
Q

For capacitors in series, what formula can be used to calculate the total capacitance of a circuit?

A

1/Ctotal = 1/C1 + 1/C2 + …

Where:

Ctotal = total capacitance (F)
C1 = capacitance of first capacitor (F)
C2 = capacitance of second capacitor (F)
… = capacitance of additional capacitors, similar to above

46
Q

What is the total capacitance in the circuit below?

A

Ctotal = 2F

1/Ctotal = 1/C1 + 1/C2
1/4 + 1/4 = 2/4 = 1/2
Ctotal = 2F

47
Q

What is a dielectric material?

A

It is an electrical insulator.
It decreases the electric field produced by a given charge.

All materials have a dielectric constant, k, which defines how effective that material is in decreasing the electric field. The dielectric constant is always equal to or greater than 1.

48
Q

Water has a dielectric constant of 80.4, while glycerine has a dielectric constant of 42.5. Which of these two substances is better able to reduce a charge’s electric field?

A

water

Remember that a higher k value indicates that a substance is more effective at insulating a charge, which lessens the field produced by that charge.

49
Q

What is the formula for capacitance between parallel plates?

A

C=kε0A / d

Where:

C = capacitance (F)
A = area of overlap of the two plates (m2)
k = dielectric constant of the material between the plates; note that for a vacuum, k = 1
ε0 = permittivity constant (ε0 = 8.85×10−12 F/m)
d = distance between the plates (m)

50
Q

How will capacitance change if a dielectric is inserted between the parallel plates of a capacitor?

A

Capacitance increases proportionally to k.

C = kε0A / d
Note that k is never less than 1, and the term “dielectric” generally denotes a k value that is somewhat greater than 1. So, adding a dielectric must increase capacitance.

51
Q

When a dielectric is inserted between the plates of a capacitor, the charge on those plates doubles. What happens to the voltage?

A

It remains the same.

In a fixed circuit, the voltage provided by a battery is always constant.

Specifically, capacitance of the circuit doubles proportionally to the charge. According to the equation C = Q / V, if both C and Q double, voltage will not change.

52
Q

What formula can be used to calculate the energy stored in a charged capacitor?

A

E = (1/2)CV2

Where:

C = capacitance (F)
V = voltage of the circuit (V)

Note that the energy stored in a capacitor will be equal to the work done to charge it initially.

53
Q

How much energy is required to charge a 10 F capacitor on a 6 V circuit?

A

180 J

E = (1/2)CV2

In this example, C = 10 F, while V = 6 V.
E = (1/2)(10)(62)
(1/2)(10)(36) = 180 J

54
Q

When a dielectric is inserted between the plates of a capacitor, the charge on those plates doubles. What happens to the capacitance?

A

Capacitance will double.

C = Q/V. Since adding a dielectric does not affect the voltage, Q and C will undergo proportional changes. Doubling Q will cause C to double as well.

55
Q

Define:

conductivity

A

It is the measure of how easily electric current flows through a material. It is represented by σ (sigma) or κ (kappa).

Conductivity is the inverse of resistivity and has units of S/m, where S refers to the siemen, an SI unit.

56
Q

The resistivity of iron is 10-7 Ωm. What is its conductivity?

A

107 S/m

Since conductivity is the reciprocal of resistivity, conductivity = 1/10-7 = 107.

57
Q

What quantity is generally measured in watts?

A

power

Power is a measure of work done over time.

1 watt is equal to 1 J/s. For the MCAT, be sure to remember that joules are the units for both work and energy.

58
Q

Define:

power

A

It is the rate at which energy is transferred per unit time.

In other words, power is equal to work over time.

59
Q

What equation would you use to find the amount of energy dissipated by a resistor over a certain amount of time?

A

Since this question deals with a resistor, it would be easiest to use P = I2R.
Alternatively, P = IV is also a valid equation, and is derived from Ohm’s Law.

I = current in A, V = voltage in V, and R = resistance in Ω.

60
Q

In a certain circuit, the voltage is doubled while the current is halved. How will the amount of power required to transfer charge change?

A

Power required does not change.

P0 = IV
Under the new conditions, V’ = 2V and I’ = I/2.

P’ = I’V’ = (I/2)(2V) = IV = P0

61
Q

In a certain circuit, the resistance remains constant while the current doubles. How will the amount of power dissipated by the circuit change?

A

Power dissipated will be increase by a factor of 4.

Use P = I2R. With resistance constant, doubling current will quadruple power, since power is directly proportional to the square of the current.

62
Q

What does the abbreviation RMS stand for?

A

RMS stands for root mean square. It represents the relationship between the instantaneous peak value and the average (mean) value.

RMS can be calculated by taking the peak value and dividing by √2. In the image above, the peak current is 1A, so the RMS current will be 1/√2 or .7A.

63
Q

Define:

AC current

A

AC current is alternating current. It periodically reverses direction of flow, creating both positive and negative values. This differs from DC (direct current), which flows in only one direction and always remains positive.

To convert AC into DC, divide the peak AC current by √2.

64
Q

How can power be calculated in an AC system?

A

AC power can be calculated in a similar way to DC power. The only difference is that all equation variables must be RMS equivalents.

PRMS = IRMSVRMS
PRMS = I2RMSR

Note that, to make any AC value into a DC equivalent, divide the peak AC by √2.

65
Q

The voltage of an AC circuit is halved while the current doubles. How will the amount of power required to transfer charge change?

A

Power will remain the same as in the original circuit.

PRMS = VRMS*IRMS
Under the new conditions, V’ = V/2 and I’ = 2I.

P’ = V’I’ = (V/2)(2I) = VI = PRMS

66
Q

Which circuit component can either break a circuit or change between distinct circuit pathways?

A

An electrical switch can break or alter a circuit.

Be careful to note whether switches on the MCAT are open or closed. Open switches prevent current from traveling in that direction, while closed switches allow current to pass through.

67
Q

In the circuit shown below, how much current is flowing through the 10Ω resistor?

A

1A

Since the switch is open in this circuit, no current flows through either the 4Ω or the 6Ω resistor. This means that the 10Ω resistor is the only one allowing current to pass through.

I = V/R = 10/10 = 1A

68
Q

What voltage is required for 4A of current to be flowing through the circuit below?

A

24V

First, find total resistance.

1/Rtotal = 1/R1 + 1/R2
1/8 + 1/24 = 4/24 = 1/6
Rtotal = 6Ω

Then, use Ohm’s law to find voltage.

V = IR = (4)(6) = 24V

69
Q

In the circuit below, the total current is 1 A. If the 8Ω and 6Ω resistors were removed while the battery is left unchanged, by what factor will the current be altered?

A

The new current will be increased by a factor of 8.

First, find the original resistance.
R = 8 + 1 / (1/4+1/4) + 6
8 + 2 + 6 = 16 Ω
I = V/R = V/16

The new resistance is 1 / (1/4+1/4) = 2 Ω.
I’ = V/R’ = V/2.

Finally, comparing the original I to the new I. The ratio is 1:8.

MCAT Physics (11 decks)

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