Electrostatics, Magnetism & Circuits

Question 1

What unit measures charge, and what signs can be given to it?

Answer 1

The Coulomb, C, is the SI unit of charge and can have values that are positive (for current or protons) or negative (for electrons).

Since the AP Physics exam is concerned primarily with current (+) moving, Coulombs will primarily be positive values.

Question 2

What are the rules for attraction of charges?

Answer 2

Opposite charges (+,-) are attracted towards each other.

The force of attraction varies proportionally with magnitude of charges.

Question 3

What are the rules for repulsion of charges?

Answer 3

Like charges (+,+ or -,-) are repulsed away from each other.

The force of repulsion varies proportionally with magnitude of charges.

Question 4

What must be true of the total charge in a closed system?

Answer 4

Total charge in a closed system will stay constant. This is the law of conservation of charge.

Question 5

Describe the total charge before and after in the following closed system: 5 protons and 4 electrons collide with sufficient energy to create 4 neutrons and 1 proton.

Answer 5

Charge remains constant at +1.

Charge initially is: +5 + (-4) = +1.

In a closed system charge must be conserved. The final charge confirms this: 4(0) + (+1) = +1.

Question 6

Define:

conductance of a material

Answer 6

Conductance is the ability of a material to transfer charge.

Conductance is generally higher in metals and lower in nonmetals.

Question 7

Define:

a conductor

Answer 7

A conductor contains many electrical charges within the medium, and they are relatively moveable.

Common conductors are usually metals with high atomic weight, such as silver, copper, and gold.

Question 8

Calcium contains 3x more moveable electric charges than Iron. Which is a better conductor?

Answer 8

Calcium has 3x the conductance of iron.

The more movable charges a material has, the higher its relative conductance.

Question 9

Define:

insulation

Answer 9

Insulation is the degree to which a material is unable to transfer charge.

Insulation is generally higher in polymers, amorphous crystals, or ionic solids. Insulation is low in all other substances.

Question 10

Define:

an insulator

Answer 10

An insulator has few free electrical charges within the medium, and those are difficult to move.

Common insulators are as glass, quartz, rubber and teflon.

Question 11

Rubber has 10,000x more moveable electric charges than paraffin. Which is a better insulator?

Answer 11

Paraffin has 10,000x higher resistance (insulation) than rubber.

The fewer moveable charges a material has, the greater its relative resistance (insulation).

Question 12

What is the relationship for the force between two charged particles separated at a given distance?

Answer 12

Coulomb’s Law, defined as:

F = Kq₁q₂ / r²

Where:
K = Coulombic Constant in Nm²/C²
q₁ and q₂ = charges in C
r = distance between the charges in m

Question 13

What does a positive value of F indicate, in Coulomb’s Law for electrostatic force?

Answer 13

A positive value of F indicates that the charges are the same sign, and will experience a force of repulsion.

Question 14

What does a negative value of F indicate, in Coulomb’s Law for electrostatic force?

Answer 14

A negative value of F indicates that the charges are opposite in sign, and will experience a force of attraction.

Question 15

What will be the change in force, if two positive charges separated by a distance r are now moved to a distance 2r apart?

Answer 15

The charges will now experience 1/4 original force.

From: F = Kq₁q₂ / r² the force is inversely proportional to the square of distance. Doubling distance will reduce magnitude of force to 1/4 its original value.

Question 16

What will be the change in force, if a negative and positive charge at a distance r are replaced by equivalent but now both negative charges?

Answer 16

Force will change from negative (attraction) to positive (repulsion).

Magnitude will be the same, since the strength of the charges remains the same.

Question 17

What concept and direction describes the path of motion that a positive test charge will travel, when put next to a stationary positive charge?

Answer 17

Electrostatic field lines.

By convention, all positive charges have field lines with arrows pointing outwards, since a positive test charge will be repulsed.

Question 18

What concept and direction describes the path of motion that a positive test charge will travel, when put next to a stationary negative charge?

Answer 18

Electrostatic field lines.

By convention, all negative charges have field lines with arrows pointing inwards, since a positive test charge will be attracted.

Question 19

Describe the motion of a positive test charge, if it’s placed exactly at the midpoint between two equal positive stationary charges.

Answer 19

It will remain immobile.

The two positive charges repel the test positive charge equally, hence cancelling each other’s force vector. There is no net force on the test charge.

Question 20

What direction will a positive charge move, when introduced into a uniform electric field pointing directly upwards?

Answer 20

Directly upwards.

A positive charge always follows the electric field.

Question 21

What direction will a negative charge move, when introduced into a uniform electric field pointing directly upwards?

Answer 21

Directly downwards.

A negative charge always goes exactly opposite to the field lines.

Question 22

What direction will a non-charged particle move, when introduced into a uniform electric field pointing directly upwards?

Answer 22

No motion. A non-charged particle (neutral) will not be accelerated in any direction by an electric field.

Question 23

What direction will a proton move, when introduced into a uniform electric field pointing towards the right?

Answer 23

To the right.

Positive charges always move in the direction of field lines.

Question 24

What direction will an electron move, when introduced into a uniform electric field pointing towards the left?

Answer 24

To the right.

Negative charges move in the exact opposite direction of field lines.

Question 25

Define:

electric potential

Answer 25

Electric potential (or electric potential energy) is the energy of position for charges, relative to each other.

This is the same as the electrostatic force between charges times the distance between them. Electric potential may be positive or negative, depending on the sign of the charges.

Question 26

What is the relationship between charges and distance that gives electric potential?

Answer 26

U = Kq₁q₂ / r

Potential energy, expressed in science as U
Where:
K = Coulombic constant in Nm²/C²
q₁ and q₂ = charges in C
r = distance between the charges in m

Question 27

If the distance between two positive charges doubles, what will happen to the electic potential between them?

Answer 27

Electric potential will halve.

From: U = Kq₁q₂ / r
U is inversely proportional to r, hence doubling r will halve U.

Question 28

Define:

potential difference between two points

Answer 28

Potential difference, also called voltage, is the potential energy per charge that is lost or gained when a test charge is moved between two positions.

By convention, the moving charge is assumed to be a positive unit test charge of 1 Coulomb.

Question 29

What is the relationship between charge and distance that gives potential difference (voltage)?

Answer 29

V = K*q / r

Where:
K = Coulombic constant in Nm²/C²
q = static charge in C
r = distance between the charge and a positive unit test charge in m

Question 30

What change in voltage is necessary to hold a unit charge in position, if the main charge is suddenly doubled?

Answer 30

Voltage must double.

From: V = K*q / r
Voltage and main charge are directly proportional, hence doubling one will double the other.

Question 31

Define:

absolute potential of a point in space

Answer 31

The absolute potential of a point in space is the voltage necessary to bring a test charge from infinity to that point, a given distance away from a fixed charge.

By convention, the test charge is assumed to be a positive unit charge of 1 Coulomb. This value can be positive or negative depending on the main charge.

Question 32

Given the absolute potential for a fixed positive charge at a known position, what would be the absolute potential for a negative charge of the same magnitude at the same position?

Answer 32

The negative of the absolute potential value that was given for the positive charge.

Absolute potential will be positive or negative depending on the sign of the charge.

Question 33

What concept describes the set of positions of equal energy that a positive test charge can be placed, when put next to a stationary charge?

Answer 33

Equipotential lines.

All charges have equipotential lines that form closed, non-overlapping loops around them.

Question 34

A solitary 5C charge is fixed in space and a test charge is placed at 1cm and then 2cm away. Can these two positions be on the same equipotential line?

Answer 34

No. These are not positions where the potential electrostatic energy is constant.

From: energy = Kq₁q₂ / r , the potential energy is inversely proportional to the distance between the fixed charge and the test charge. Doubling the distance will halve the energy, hence these positions cannot be equal energy and cannot be equipotential.

Question 35

Define:

an electric dipole

Answer 35

An electric dipole occurs as the result of any distinct separation of positive and negative charge in an object or system.

On the AP Physics exam, the simplest case is always assumed: a linear object with one positive and one negative end.

Question 36

Describe the motion of a dipole placed into an electric field.

Answer 36

The dipole will rotate so that the positive end is aligned with the electric field and the negative end is exactly opposite to the field.

Assuming that the dipole is overall neutral, there will not be any translation in the field.

Question 37

Define:

electric current, I

Answer 37

Electric current, I, is the directional flow of charge through a conducting medium.

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

Question 38

What is the formula and SI unit for current?

Answer 38

Current: I = ∆Q / ∆t

Where:
∆Q = change in charge (coulombs)
∆t = change in time (seconds)

The SI unit of charge is the ampere, A, and represents 1 C/s.

Question 39

If electrons are moving along a wire from left to right, where is conventional current moving?

Answer 39

Conventional current is moving from right to left.

Conventional current always refers to the flow of positive charge, and will always be opposite to the flow of electrons.

Question 40

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

Answer 40

The current will be 4x larger.

I₀ = Q / t
given new Q’ = 2Q
and new t’ = (1/2)t
then: I’ = Q’ / t’
= 2Q / (1/2)t = 4(Q / t)
= 4I₀

Question 41

Define:

a battery

Answer 41

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 undergoing redox chemical reactions in the cell.

Question 42

Define:

electromotive force (EMF)

Answer 42

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

This is not actually a “force”. EMF is an energy per charge, i.e. work done to move charge (Volts).

Question 43

Define:

electric potential energy

Answer 43

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 AP Physics exam.

Question 44

What is the formula and SI unit of electric potential energy?

Answer 44

U = kQq/r

Where:
k = Coulomb’s constant (9x10⁹ Nm²/C²)
Q = primary charge in C
q = test (secondary) charge in C
r = radius between charges in m

As with all forms of energy, the SI unit is the Joule, J, and represents 1 N*m.

Question 45

What does a positive vs. a negative electric potential energy indicate about the charges and force between them?

Answer 45

A positive energy happens when both charges are the same parity (+,+ or –,–), and will cause a repulsive force between the two charges.

A negative energy happens when both charges are opposite parity, and will cause an attractive force between the two charges.

Question 46

Define:

voltage

Answer 46

Voltage (electric potential) is the electrical energy per unit charge necessary to move a test charge against the electric field of a fixed charge.

Question 47

What is the formula and SI unit of voltage?

Answer 47

V = kQ/r

Where:
k = Coulomb’s constant (9x10⁹ Nm²/C²)
Q = primary charge in C
r = radius between charges in m

Voltage is in SI units of volts, V, and represents 1 J/C.

Question 48

What is the change in voltage if a test charge at a distance r from a central charge Q is now moved twice as far away?

Answer 48

The voltage will decrease by 1/2.

From: V₀=kQ/r
new r’=2r
k and Q are constant

V’‘=kQ/r’=kQ/2r
=(kQ/r)(1/2)=V₀/2

Question 49

Define:

resistance

Answer 49

Resistance is a measure of how difficult it is to pass current through some object. Many factors affect resistance, including: cross-sectional area, length, and general resistivity.

Though all materials have some resistance at room temperature, super-cooled superconductors have a resistance of zero.

Question 50

What is the formula for calculating the total resistance of an object, and what are the SI units of resistance?

Answer 50

R = ρ***L / A

Where:
ρ = resistivity of the material in Ωm
L = length of the object along current in m
A = cross sectional area in m²

Resistance is in SI units of Ohms, Ω, and represents 1 J*s/C².

Question 51

By what factor has total resistance changed if the length of a resistor is doubled?

Answer 51

The total resistance will also double.

From: R₀ = ρ*L/A
new L’ = 2L

R’ = ρ*L’/A
= ρ*2L/A = 2(ρ*L/A)=2R₀

Question 52

By what factor has total resistance changed if the cross-sectional area of a resistor is doubled?

Answer 52

The total resistance will decrease by 1/2.

From: R₀ = ρ*L/A
new A’ = 2A

R’ = ρ*L/A’
= ρ*L/2A = (ρ*L/A)(1/2)=R₀/2

Question 53

By what factor has total resistance changed if the length of a resistor doubles and the cross sectional area also doubles?

Answer 53

The total resistance will remain constant.

From: R₀ = ρ*L/A
new L’ = 2L
new A’ = 2A

R’ = ρ*L’/A’
= ρ*2L/2A = ρ*L/A=R₀

Question 54

By what factor has total resistance changed of a wire, if the length doubles and the radius also doubles?

Answer 54

The total resistance will decrease by 1/2.

From: R₀ = ρ*L/A
new L’ = 2L
new A’ = 4A (since area goes as the square of the radius)

R’ = ρ*L’/A’
= ρ*2L/4A = ρ*L/2A=R₀/2

Question 55

Define:

Internal resistance of a battery

Answer 55

Internal resistance of a battery is the difference between the predicted ideal voltage that a battery should deliver vs. the real-world value that it provides.

Occasionally, internal resistance may be referred to as the “impedence”, since factors like battery size, chemical makeup, and current load can all impede ideal voltage.

Question 56

Define and give the SI units for:

Resistivity

Answer 56

Resistivity, ρ, is the measurement of how much a certain material resists the flow of current, and has SI units of Ω*m.

Question 57

What is the formula for Ohm’s law?

Answer 57

Ohm’s law is written as:

I = V/R
or commonly: V = I*R

Where:
V = voltage in V
I = current in A
R = resistance in Ω

Question 58

What must happen to the resistance, if the voltage is doubled in a circuit that has constant current flow?

Answer 58

The resistance must double also.

From Ohm’s law: V=IR

Since current is held constant, voltage and resistance are directly proportional. Doubling voltage must result in resistance doubling.

Question 59

What must have happened to the voltage supplied to a fixed resistance circuit, if the current suddenly drops to half original?

Answer 59

The voltage must have also dropped to half.

From Ohm’s law: V=IR

Since R is held constant, V and I are directly proportional. Halving voltage means current must be halved also.

Question 60

What must have happened to the resistance in a fixed voltage circuit, if the current suddenly doubles?

Answer 60

The resistance must have dropped to 1/2 original.

From Ohm’s law: V=IR

Since V is held constant, R and I are inversely proportional. Doubling current means resistance must be halved.

Question 61

What does this symbol stand for in a circuit diagram?

Answer 61

This is a battery. Batteries will always have a larger positive terminal (labeled +) and a smaller negative terminal (labeled –).

Conventional current always flows from + to – around the circuit.

Question 62

What does this symbol stand for in a circuit diagram?

Answer 62

This is a resistor. Resistors impede current flow and have units of ohms.

Question 63

What does this symbol stand for in a circuit diagram?

Answer 63

This is a capacitor. Capacitors store charge and have units of farads.

Question 64

What does this symbol stand for in a circuit diagram?

Answer 64

This is a switch. Switches can be open (as shown) so that no current flows around the circuit, or closed so that both dots are connected and current flows.

Question 65

What does this symbol stand for in a circuit diagram?

Answer 65

This is the symbol for ground. 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 ground.

Question 66

What is the difference visually between resistors in parallel vs. series in a circuit?

Answer 66

Resistors in parallel are each on a diverging circuit pathway. A single charge will pass through either one or the other to complete the circuit. (first image)

Resistors in series follow each other on continuous circuit pathway. A single charge must go through both to complete the circuit. (second image)

Question 67

What is the formula for calculating the total resistance of a circuit, for resistors in parallel?

Answer 67

1/R_Total = 1/R₁ + 1/R₂ + …

Where:
R_Total = total resistance in Ω
R₁ = first resistor in Ω
R₂ = second resistor in Ω
… = any additional resistors in similar fashion

Question 68

What is the total resistance in the circuit below?

Answer 68

R_Total = 5 Ω

1/R_Total = 1/R₁ + 1/R₂
= 1/10 + 1/10 = 2/10
= 1/5

Question 69

What is the formula for calculating the total resistance of a circuit, for resistors in series?

Answer 69

R_Total = R₁ + R₂ + …

Where:
R_Total = total resistance in Ω
R₁ = first resistor in Ω
R₂ = second resistor in Ω
… = any additional resistors in similar fashion

Question 70

What is the total resistance in the circuit below?

Answer 70

R_Total = 10 Ω

R_Total = R₁ + R₂
= 4 + 6 = 10

Question 71

What is the total resistance in the circuit below?

Answer 71

The total resistance is 5Ω.

First find resistance for the 4Ω and 6Ω in series.
R_series = 4 + 6 = 10Ω.

Then find resistance for the 10Ω and 10Ω in parallel.
1/R_Total = 1/10 + 1/10
= 2/10 = 1/5.
R_Total = 5Ω.

Question 72

What amount of current must be flowing around the circuit below?

Answer 72

The current is 2A.

First find resistance for the 4Ω and 6Ω in series.
R_series = 4 + 6 = 10Ω.

Then find resistance for the 10Ω and 10Ω in parallel.
1/R_Total = 1/10 + 1/10
= 2/10 = 1/5.
R_Total = 5Ω.

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

Question 73

What amount of current must be flowing around the circuit below?

Answer 73

The current is 1A.

First find resistance for the 12Ω and 6Ω in parallel.

1/R_Total = 1/12 + 1/6
= 3/12 = 1/4.
R_Total = 4Ω.

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

Question 74

Define:

Capacitance

Answer 74

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

Generally capacitance implies “parallel plate capacitance”, where charge is related to the area of the plates, distance between plates, and delectric material between plates.

Question 75

What is the formula and SI unit for capacitance?

Answer 75

Capacitance: C = Q / V

Where:
C = capacitance
Q = charge transferred
V = voltage

Capacitance is in SI units of farads, F, and represents 1 C/V.

Question 76

What has happened to the capacitance, if the charge on a capacitor plate doubles while the voltage remains constant?

Answer 76

Capacitance will also double.

From capacitance: C₀ = Q / V
new Q’ = 2Q

C’ = Q’/V = 2Q/V = 2(Q/V)= 2C₀

Question 77

What is the formula for calculating the total capacitance of a circuit, for capacitors in parallel?

Answer 77

C_Total = C₁ + C₂ + …

Where:
C_Total = total capacitance in F
C₁ = first capacitor in F
C₂ = second capacitor in F
… = any additional capacitors in similar fashion

Question 78

What is the total capacitance in the circuit below?

Answer 78

Capacitance is 10F.

C_Total = C₁ + C₂
= 8 + 2 = 10F

Question 79

What is the formula for calculating the total capacitance of a circuit, for capacitors in series?

Answer 79

1/C_Total = 1/C₁ + 1/C₂ + …

Where:
C_Total = total capacitance in F
C₁ = first capacitor in F
C₂ = second capacitor in F
… = any additional capacitors in similar fashion

Question 80

What is the total capacitance in the circuit below?

Answer 80

Capacitance is 2F.

1/C_Total = 1/C₁ + 1/C₂
= 1/4 + 1/4 = 2/4 = 1/2.
C_Total = 2F.

Question 81

Define:

a dielectric material

Answer 81

A dielectric material is an electrical insulator, and 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.

Question 82

Water has a dielectric constant k = 80.4 and glycerine has a dielectric constant of k = 42.5. Which of these two substances would lessen a charge’s electric field more?

Answer 82

Water is a better dielectric than glycerin.

Recall: a higher k value indicates that substance is more effective at insulating a charge and lessening a charge’s field.

Question 83

What is the formula for capacitance between parallel plates?

Answer 83

C=k*ε₀*A / d

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

Question 84

What will happen to capacitance, if a dielectric with k>1 is inserted between the parallel plates of a capacitor?

Answer 84

Capacitance increases proportional to k.

Since: C=k*ε₀*A / d
and for air k=~1
any k value greater than 1 will directly increase capacitance.

Question 85

What has happened to the voltage of a fixed circuit, if a dielectric material is inserted between plates of a capacitor and the charge on those plates doubles in order for current to continue to flow?

Answer 85

Voltage provided by a battery is always constant in a fixed circuit.

Capacitance of the circuit will have increased; but since current is still flowing, it must be true that nothing has been done to change the voltage of the battery.

Question 86

What is the formula for calculating the energy stored in a charged capacitor?

Answer 86

E = (1/2)CV²

Where:
C = capacitance in farads
V = voltage of the circuit in volts

Note: energy stored in a capacitor is necessarily the same as the work done initially to charge it.

Question 87

What is the energy necessary to charge a 10F capacitor on a 6V circuit?

Answer 87

180 J

From E = (1/2)CV²

C = 10F, V = 6V.
E = (1/2)*10*6²
= (1/2)*10*36=180J.

Question 88

What has happened to the capacitance of a fixed voltage circuit, if a dielectric material is inserted between plates of a capacitor and the charge on those plates doubles in order for current to continue to flow?

Answer 88

Capacitance will double.

From the formula: C = Q/V

Since V is held constant, Q and C are directly proportional. Doubling charge must mean that capacitance doubles also.

Question 89

Define:

Conductivity

Answer 89

Conductivity is the measure of how well a material is able to conduct electric current through it, and is represented by σ (sigma) or κ (kappa).

Conductivity is the inverse of resistivity and will have units of S/m, where S is the SI unit of siemens.

Question 90

What must the conductivity of iron be, if it is found that the resistivity of iron is 10^-7Ω*m?

Answer 90

10⁷ S/m

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

Question 91

What must have happened to the conductivity of a metal, if raising the temperature halves the resistance?

Answer 91

The conductivity increases to double.

Conductance and resistance are inversely proportional, so halving one will double the other.

Question 92

Define and give the SI units for:

Power

Answer 92

Power is the rate at which energy is transferred per unit time and has SI units of Watts (W), which represents J/s.

Question 93

What are the formulas for power transferred through a circuit and power dissipated by a circuit?

Answer 93

Power transferred is P = IV

Power dissipated is P = I²R

Where:
I = current in A
V = voltage in V
R = resistance in Ω

Question 94

How will the amount of power required to transfer charges change, if the voltage of the circuit doubles while the current halves?

Answer 94

Power required is the same.

From P₀ = IV
V’ = 2V, I’ = I/2

P’ = I’V’ = (I/2)(2V) = IV = P₀

Question 95

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

Answer 95

Power dissipated will be 4x original.

From P = I²R

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

Question 96

Define:

RMS

Answer 96

RMS stands for Root Mean Square, and 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 below, the peak current is 1A, so the RMS current will be 1/√2 = .7A

Question 97

Define:

AC current

Answer 97

AC current (alternating current) periodically reverses direction of flow, creating both positive and negative values.

This is different from DC (direct current), which flows in only one direction and remains always positive.

Note: to make an AC current value into a DC current equivalent, divide the peak AC current by √2, this is known as the RMS factor.

Question 98

Define:

AC Voltage

Answer 98

AC voltage (alternating voltage) periodically reverses sign, having both positive and negative values, due to AC current changing direction and sign.

This is different from normal voltage (direct voltage), which remains always positive, due to supposing that current flows in one direction.

Note: to make an AC voltage value into a DC voltage equivalent, divide the peak AC voltage by √2, this is known as the RMS factor.

Question 99

Define and give the formula for:

AC Power

Answer 99

AC power (alternating power) is the energy transferred by an alternating current.

This is similar to normal power, the only difference is that all equation variables are the RMS equivalents.

P_RMS = I_RMS*V_RMS
P_RMS = I²_RMS*R

Question 100

How will the amount of power required to transfer charges change, if the voltage of an AC circuit halves while the current doubles?

Answer 100

Power will remain the same as original.

From P_RMS = V_RMS*I_RMS
V’ = V/2, I’ = 2I

P’ = V’I’ = (V/2)*2I = VI = P_RMS

Question 101

Define:

an electrical circuit switch

Answer 101

An electrical switch is a circuit component that can break current pathways, or change between pathways.

Question 102

What is the amount of current flowing through the 10Ω resistor is the circuit as currently shown below?

Answer 102

Current is 1A through the 10Ω resistor.

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

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

Question 103

What voltage battery is necessary in order for 4A of current to be flowing through the circuit below?

Answer 103

Voltage must be 24V.

First, find total resistance:

1/R_Total = 1/R₁ + 1/R₂
= 1/8 + 1/24 = 4/24
= 1/6.
R_Total = 6Ω

Then, From Ohm’s law:
V=IR = 4*6 = 24V.

Question 104

In the circuit below, the total amount of current flowing is 1A. If both the 8Ω and 6Ω resistors were removed and the same battery used, by what factor will the current be changed?

Answer 104

The new current will be 8x increased.

Original resistance was
R = 8 + 1/(1/4+1/4) + 6
= 8 + 2 + 6 = 16Ω.
I = V/R = V/16.

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

Comparing I : I’ = 1:8.