Current, Voltage, and Resistance

The video lesson for this topic is coming soon.

Essential Question

What makes charges flow through a circuit?

A circuit needs a closed loop and a battery, but how much charge flows, and how hard is it pushed? By the end of this lesson, you should be able to explain current, voltage, and resistance, and solve problems with I = q/t and ΔV = PE/q.

Electric Current

Current: the flow of charge
Amount of charge per second
Unit: ampere (A)

A section of wire drawn as a gray tube with blue dots for charges moving through it. A dashed line across the tube is labeled count here. A red arrow below the tube is labeled current I. Caption: 1 A means 1 coulomb passes each second.

Current is the rate at which charge flows past a point in a circuit. Imagine standing at one spot in the wire and counting how many coulombs of charge go by each second. Current is measured in amperes (A), often shortened to amps. One ampere means one coulomb of charge passes each second. By convention, current points the way positive charge would move: out of the + terminal of the battery.

The Current Equation

I = q / t

I = current, in amperes (A)
q = charge, in coulombs (C)
t = time, in seconds (s)

The equation I = q / t in a yellow box. Below it, three cards: I is current in amperes, q is charge in coulombs, and t is time in seconds.

Dividing coulombs by seconds gives amperes, so 1 A = 1 C/s. The equation can be rearranged to q = It to find charge, or t = q/I to find time.

Voltage

Energy given to each coulomb
The “push” from the battery
Unit: volt (V) = 1 J/C

Top: a 1.5 V cell with an arrow to a card that says each coulomb gets 1.5 J. Bottom: a 9 V battery with an arrow to a card that says each coulomb gets 9 J. Caption: more volts, more energy per coulomb.

Voltage is the energy the battery gives to each coulomb of charge. It is the “push” that makes charge flow. Voltage is measured in volts (V), and one volt means one joule of energy per coulomb. A 1.5 V cell gives 1.5 J to every coulomb that passes through it, and a 9 V battery gives 9 J to every coulomb. Voltage is also called potential difference, which is why its symbol is ΔV.

The Voltage Equation

ΔV = PE / q

ΔV = voltage (potential difference), in volts (V)
PE = electric potential energy, in joules (J)
q = charge, in coulombs (C)

The equation delta V = PE / q in a yellow box. Below it, three cards: delta V is voltage in volts, PE is electric potential energy in joules, and q is charge in coulombs.

Dividing joules by coulombs gives volts, so 1 V = 1 J/C. Rearranged, PE = qΔV gives the energy a charge picks up from a battery.

Resistance

Opposes the flow of charge
Unit: ohm (Ω)
More resistance, less current
Energy turns into heat

A block of material with two rows of gray atoms. A blue zigzag path shows a charge bouncing back and forth off the atoms as it moves left to right. Caption: charges bump into atoms, which heats the material.

Resistance is how strongly a material opposes the flow of charge. As charges move through a material, they keep bumping into its atoms, which slows them down and turns some of their energy into heat. Resistance is measured in ohms (Ω). For the same voltage, more resistance means less current. A resistor is a part built to have a certain resistance, and a bulb’s filament is a resistor that gets hot enough to glow.

What Affects Resistance?

Longer wire: more resistance
Thicker wire: less resistance
Material matters

Three comparisons of wires. A short wire has less resistance than a long wire. A thin wire has more resistance than a thick wire. A copper wire has less resistance than a nichrome wire of the same size.

Three things control a wire’s resistance. A longer wire has more resistance, because charges bump into more atoms along the way. A thicker wire has less resistance, because it has more room for charge to flow, like a wider hallway. The material matters too: copper has very low resistance, so it is used for wires, while nichrome has high resistance, so it is used in toasters and heaters.

The Water Analogy

Pump ↔ battery
Water flow ↔ current
Pump’s push ↔ voltage
Narrow pipe ↔ resistance

A closed loop of water pipe. A pump on the left side pushes water around the loop, shown by blue arrows. On the right side, part of the pipe is narrow. Caption: the pump pushes, the narrow pipe resists, and the water flows.

A circuit works a lot like water in a closed loop of pipe. The pump is like the battery: it gives the water a push, which is like voltage. The amount of water flowing past a point each second is like current. A narrow section of pipe makes it harder for water to get through, just like a resistor. A stronger pump or a wider pipe means more flow.

Worked Examples

Try each problem on your own first, then check your work against the solution.

Example 1: Finding Current

A total charge of 12 C flows through a bulb in 4.0 s. What is the current in the bulb?

Knowns: q = 12 C, t = 4.0 s
Equation: I = q/t
Substitute: I = 12 C / 4.0 s
Answer: I = 3.0 A. A coulomb per second is an ampere.

Example 2: Finding Charge

A toaster draws a current of 2.5 A for 30.0 s. How much charge flows through it?

Knowns: I = 2.5 A, t = 30.0 s
Equation: I = q/t, so q = It
Substitute: q = (2.5 A)(30.0 s). An amp is a coulomb per second, so the seconds cancel.
Answer: q = 75 C

Example 3: Finding Voltage

A battery gives 36 J of energy to 3.0 C of charge. What is the battery’s voltage?

Knowns: PE = 36 J, q = 3.0 C
Equation: ΔV = PE/q
Substitute: ΔV = 36 J / 3.0 C
Answer: ΔV = 12 V. A joule per coulomb is a volt.

Example 4: Finding Energy

A 9.0 V battery moves 0.50 C of charge through a circuit. How much energy does the battery give to the charge?

Knowns: ΔV = 9.0 V, q = 0.50 C
Equation: ΔV = PE/q, so PE = qΔV
Substitute: PE = (0.50 C)(9.0 V). A volt is a joule per coulomb, so the coulombs cancel.
Answer: PE = 4.5 J

Big Ideas

Current is the flow of charge: I = q/t (A)
Voltage is energy per charge: ΔV = PE/q (V)
Resistance opposes current (Ω)
Long, thin wires have more resistance

Back to the essential question: a battery’s voltage gives energy to each coulomb of charge and pushes it around the loop. The flow of charge is the current, and resistance pushes back against that flow. Next up, we’ll connect all three with Ohm’s law.