Circuits

Question 1

Define:

electric current (I)

Answer 1

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.

Question 2

What formula and SI unit are associated with current?

Answer 2

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.

Question 3

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

Answer 3

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.

Question 4

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

Answer 4

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

Question 5

Define:

battery

Answer 5

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.

Question 6

Define:

electromotive force

(EMF)

Answer 6

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.

Question 7

Define:

electric potential energy

Answer 7

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.

Question 8

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

Answer 8

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.

Question 9

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

Answer 9

• 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.

Question 10

Define:

voltage

Answer 10

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.

Question 11

What formula and SI unit are associated with voltage?

Answer 11

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.

Question 12

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?

Answer 12

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

Question 13

Define:

resistance

Answer 13

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.

Question 14

What quantity is generally measured in ohms?

Answer 14

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

1 Ω represents 1 J*s / C².

Question 15

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

Answer 15

R = ρL / A

Where:

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

Question 16

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

Answer 16

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

Question 17

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

Answer 17

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

Question 18

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

Answer 18

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

Question 19

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

Answer 19

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

Question 20

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

Answer 20

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.

Question 21

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

Answer 21

• 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.

Question 22

What is the formula for Ohm’s law?

Answer 22

V = IR.

Where:

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

Question 23

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

Answer 23

The resistance also doubles.

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

Question 24

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

Answer 24

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.

Question 25

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

Answer 25

The resistance must have **dropped by one-half**. ## Footnote According to Ohm's law, V = IR. Since voltage is held constant, resistance and current are inversely proportional.

Question 26

What does this symbol stand for in a **circuit diagram**?

Answer 26

battery ## Footnote 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.

Question 27

What does this symbol stand for in a **circuit diagram**?

Answer 27

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

Question 28

What does this symbol stand for in a **circuit diagram**?

Answer 28

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

Question 29

What does this symbol stand for in a **circuit diagram**?

Answer 29

switch ## Footnote The switch shown here is open. Switches can be open to prevent current flow around the circuit or closed to allow current flow.

Question 30

What does this symbol stand for in a **circuit diagram**?

Answer 30

ground ## Footnote 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.

Question 31

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

Answer 31

* 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.

Question 32

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

Answer 32

R_total = R₁ + R₂ + ... ## Footnote Where: R_total = total resistance (Ω) R₁ = resistance of first resistor (Ω) R₂ = resistance of second resistor (Ω) ... = resistance of additional resistors, similar to above

Question 33

What is the **total resistance** in the circuit below?

Answer 33

R_total = 10 Ω ## Footnote R_total = R₁ + R₂ 4 Ω + 6 Ω = 10 Ω

Question 34

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

Answer 34

1/R_total = 1/R₁ + 1/R₂ + ... ## Footnote Where: R_total = 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

Question 35

What is the **total resistance** in the circuit below?

Answer 35

R_total = 5 Ω ## Footnote 1/Rt_otal = 1/R₁ + 1/R₂ 1/10 Ω+ 1/10 Ω = 2/10 Ω = 1/5 Ω

Question 36

What is the **total resistance** in the circuit below?

Answer 36

5Ω ## Footnote First, find the resistance for the resistors in series. R_series = 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/R_total = 1/10 + 1/10 = 2/10 = 1/5 R_total = 5 Ω

Question 37

Find the **current** flowing within the circuit pictured below.

Answer 37

1A ## Footnote First, find the resistance for the resistors in parallel. 1/R_total = 1/12 + 1/6 = 3/12 = 1/4 R_total = 4 Ω Then, from Ohm's law: I = V/R = 4/4 = 1 A

Question 38

Find the **current** flowing within the circuit pictured below.

Answer 38

2A ## Footnote First, find the resistance for the resistors in series. R_series = 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/R_total = 1/10 + 1/10 = 2/10 = 1/5 R_total = 5 Ω Finally, from Ohm's law: I = V/R = 10/5 = 2 A

Question 39

# Define: capacitance

Answer 39

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

Question 40

What quantity is generally measured in **farads**?

Answer 40

capacitance ## Footnote 1 F represents 1 C/V. Its SI unit is F.

Question 41

What formula can be used to calculate **capacitance**?

Answer 41

C = Q / V ## Footnote Where: C = capacitance (farads) Q = charge stored (coulombs) V = voltage (V)

Question 42

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

Answer 42

Capacitance will also **double**. ## Footnote C₀ = Q / V Under the new conditions, Q' = 2Q. C' = Q' / V = 2Q / V = 2(Q / V) = 2C₀

Question 43

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

Answer 43

C_total = C₁ + C₂ + ... ## Footnote Where: C_total = total capacitance (F) C₁ = capacitance of first capacitor (F) C₂ = capacitance of second capacitor (F) ... = capacitance of additional capacitors, similar to above

Question 44

What is the **total capacitance** in the circuit below?

Answer 44

C_total = 10 F ## Footnote C_total = C₁ + C₂ 8 F + 2 F = 10 F

Question 45

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

Answer 45

1/C_total = 1/C₁ + 1/C₂ + ... ## Footnote Where: C_total = total capacitance (F) C₁ = capacitance of first capacitor (F) C₂ = capacitance of second capacitor (F) ... = capacitance of additional capacitors, similar to above

Question 46

What is the **total capacitance** in the circuit below?

Answer 46

C_total = 2F ## Footnote 1/C_total = 1/C₁ + 1/C₂ 1/4 + 1/4 = 2/4 = 1/2 C_total = 2F

Question 47

What is a **dielectric material**?

Answer 47

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

Question 48

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**?

Answer 48

water ## Footnote 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.

Question 49

What is the formula for capacitance between **parallel plates**?

Answer 49

C=kε₀A / d ## Footnote 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)

Question 50

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

Answer 50

Capacitance **increases** proportionally to k. ## Footnote 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.

Question 51

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

Answer 51

It remains the **same**. ## Footnote 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.

Question 52

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

Answer 52

E = (1/2)CV² ## Footnote 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.

Question 53

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

Answer 53

180 J ## Footnote 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

Question 54

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

Answer 54

Capacitance will **double**. ## Footnote 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.

Question 55

# Define: conductivity

Answer 55

It is the measure of how easily electric current flows through a material. It is represented by σ (sigma) or κ (kappa). ## Footnote Conductivity is the inverse of resistivity and has units of S/m, where S refers to the siemen, an SI unit.

Question 56

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

Answer 56

10⁷ S/m ## Footnote Since conductivity is the reciprocal of resistivity, conductivity = 1/10^-7 = 10⁷.

Question 57

What quantity is generally measured in **watts**?

Answer 57

power ## Footnote 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.

Question 58

# Define: power

Answer 58

It is the rate at which energy is **transferred per unit time**. ## Footnote In other words, power is equal to **work over time**.

Question 59

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

Answer 59

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

Question 60

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

Answer 60

Power required does **not** change. ## Footnote P₀ = IV Under the new conditions, V' = 2V and I' = I/2. P' = I'V' = (I/2)(2V) = IV = P₀

Question 61

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

Answer 61

Power dissipated will be **increase** by a factor of **4**. ## Footnote Use P = I²R. With resistance constant, doubling current will quadruple power, since power is directly proportional to the square of the current.

Question 62

What does the abbreviation **RMS** stand for?

Answer 62

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

Question 63

# Define: AC current

Answer 63

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

Question 64

How can **power** be calculated in an **AC** system?

Answer 64

**AC power** can be calculated in a similar way to DC power. The only difference is that all equation variables must be **RMS equivalents**. ## Footnote P_RMS = I_RMSV_RMS P_RMS = I²_RMSR Note that, to make any AC value into a DC equivalent, divide the peak AC by √2.

Question 65

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

Answer 65

Power will remain the **same** as in the original circuit. ## Footnote P_RMS = V_RMS\*I_RMS Under the new conditions, V' = V/2 and I' = 2I. P' = V'I' = (V/2)(2I) = VI = P_RMS

Question 66

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

Answer 66

An **electrical switch** can break or alter a circuit. ## Footnote 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.

Question 67

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

Answer 67

1A ## Footnote 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

Question 68

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

Answer 68

24V ## Footnote First, find total resistance. 1/R_total = 1/R₁ + 1/R₂ 1/8 + 1/24 = 4/24 = 1/6 R_total = 6Ω Then, use Ohm's law to find voltage. V = IR = (4)(6) = 24V

Question 69

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?

Answer 69

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