Units
State the units name and symbol for current
Amperes (Amps) - A
Describe what happens to current in a series circuit.
The current is the same everywhere in the circuit
A light bulb has a power rating of 60W. What does this mean?
It will draw 60 joules of energy each second.
A kettle uses 2000 J of electrical energy to heat water, but only 1600 J actually goes into raising the water's temperature. What is the efficiency of the kettle?
Efficiency = (useful energy output ÷ total energy input) × 100 = (1600 ÷ 2000) × 100 = 80%
In a series circuit with one battery and two bulbs, the current is 0.5 A at one point. What is the current at any other point in the circuit? Explain why.
The current is 0.5 A everywhere in the circuit. In a series circuit there's only one path for charge to flow, so the same amount of charge passes every point per second, current isn't "used up," it's the same throughout the loop.
State the units for resistance
Using an equation, explain what happens to voltage in a series circuit with 2 identical bulbs.
V = I x R meaning voltage is split evenly between the components. Each component has the same current flowing through them and the same resistance. The supply voltage must be evenly distributed.
What is the relationship between electrical energy and power. Describe or state the equation relationship.
Power is the rate of energy transfer. (Joules per second)
Power = ΔEnergy/Time (s). More energy transferred in the same time, more power.
Two electric motors both lift the same weight to the same height. Motor A takes in 500 J and does 350 J of useful work. Motor B takes in 400 J and does 300 J of useful work. Which motor is more efficient, and what does that tell you about the two motors that comparing their useful work output alone wouldn't?
Motor A: (350÷500)×100 = 70%.
Motor B: (300÷400)×100 = 75%.
Motor B is more efficient, even though it produces less useful work overall — efficiency measures how well a machine converts energy, not how much work it does in absolute terms. A less powerful machine can still waste proportionally less energy.
Three identical resistors are connected in series across a 12 V battery. What is the voltage across each resistor, and why does the total voltage "share out" this way rather than each resistor getting the full 12 V?
Each resistor gets 4 V (12 / 3). The total voltage supplied by the battery is shared between components in series because the energy given to the charge by the battery is used up (transferred) as it passes through each resistor in turn, by the time it's gone through all three, all the energy has been transferred, so the voltages must add up to the total.
What would the unit be for the quantity calculated by current x voltage?
Power - Watts (W)
The overall resistance will increase. As I = V/R and voltage is the same, the current will be reduced. If the voltage across the bulb remains the same, but current is reduced, it will draw less power than before.
A 24V battery is used to power a circuit containing a component with 6Ω of resistance. How much energy will be drawn in 2 minutes by the component.
V/R=I; 24 / 6 = 4A
P = IV; 4 x 24 = 96W
96 x (60 x2) = 11,520 J
A student says "a machine that is 100% efficient would be the ultimate energy-saving invention." Using what you know about energy transformations, explain why no real machine can reach 100% efficiency, and where the "lost" energy actually goes.
No machine is 100% efficient because energy is always transformed into some non-useful form during a transfer, usually heat, caused by friction, resistance, or air resistance. This "wasted" energy isn't destroyed (conservation of energy still holds); it's dispersed into the surroundings, usually as heat, and becomes harder to use for useful work.
A student adds a third bulb in series to a circuit that originally had two bulbs. She notices all the bulbs get dimmer. Using ideas about resistance and current, explain why adding another bulb affects the brightness of the others, even though nothing else in the circuit changed.
Adding a third bulb in series increases the total resistance of the circuit. Since the battery's voltage hasn't changed, but there's now more total resistance to push current through, the current throughout the whole circuit decreases (V = IR). Because all bulbs are in the same single loop, they all experience this same lower current, so they all dim together — not because the new bulb "steals" energy, but because the whole circuit now resists current flow more.
What would the unit be for the quantity found by doing RxI?
Voltage - Volts (V)
Compare the brightness between these two scenarios. You must show workings.
Circuit 1: 12V battery, 3Ω bulb and 6Ω bulb.
Circuit 2: 6V battery, two 2Ω bulbs
Circuit 1:
12/9 = 1.3A (total resistance)
V1 = 1.3 x 3 ~ 4V
V2 = 1.3 x 6 ~8V
P1 = 4 x 1.3 = 5.2W
P2 = 8 x 1.3 = 10.4W (Bulb 2 twice as bright as Bulb 1)
Circuit 2:
6/4 = 1.5A (total current)
V of bulbs = 1.5 x 2 = 3V
P of bulbs = 3 x 1.5 = 4.5W
Both bulbs are the same but less bright than circuit 1's bulbs
A device is able to draw 120W of power when connected to a 6V power supply. Calculate the resistance of the device.
P = VI; 120 = 6 x I; 120/6 = 20A
R = V/I; 6/20 = 0.3Ω
OR
P=VI and V=IR --> P = V2xR
62 / 120 = 0.3Ω
An old incandescent light bulb is about 5% efficient, while an LED bulb is about 40% efficient. If both bulbs take in 100 J of electrical energy per second, calculate how much energy is wasted per second by each bulb, and explain in energy terms (not just cost) why this matters over a year of continuous use.
Incandescent: wastes 95 J/s (95% wasted). LED: wastes 60 J/s (60% wasted). Even though both waste energy, the incandescent bulb wastes far more, this compounds significantly over a year, meaning far more total energy input is required (and more heat generated) for the same light output, which has both cost and environmental implications.
A series circuit has a 9 V battery and two resistors. The current in the circuit is 0.3 A, and the voltage across the first resistor is 3 V. Calculate the voltage across the second resistor, then calculate the resistance of each resistor.
Voltage across second resistor = 9 V − 3 V = 6 V. Resistance of first resistor: R = V/I = 3 / 0.3 = 10 Ω. Resistance of second resistor: R = 6 / 0.3 = 20 Ω.
What are the units and symbol for charge?
Coulomb (C)
Explain why voltmeters should not be connected in series despite wanting to measure the voltage of a series component.
Voltmeters measure the difference between energy/potential difference between two points, allowing us to measure how much power/voltage is being drawn by a component. If it were connected in series, it would only be measuring the potential difference of itself and there would be no difference (voltage)!
If the current of a circuit is tripled, what will happen to the power drawn by the components. Use equations to support your answer.
The power will increase. As V=IR and P=IV, tripling the current will lead to P = I2 x R, meaning power increases by a factor of 9.
A car engine is 25% efficient and a bicycle-and-rider system is roughly 20% efficient. A student claims "the car is the better machine because it has higher efficiency." Evaluate this claim — is efficiency alone enough to judge which is the "better" machine? Justify your answer using ideas beyond the efficiency number itself.
No, efficiency alone doesn't determine which machine is "better." A car and a bicycle serve different purposes, use different energy sources (fossil fuel vs human muscle), have very different power outputs and speeds, and have different environmental and practical impacts. "Better" depends on the criteria you value (speed, cost, environmental impact, health benefit, accessibility) — efficiency is only one factor among several, so the claim oversimplifies the comparison.
A student builds a series circuit with a battery, a bulb, and a resistor. They then swap the order of the bulb and resistor. Predict, with reasoning, whether the brightness of the bulb will change — and use this to explain what "series" actually means for how components share current and voltage.
The brightness will not change. In a series circuit, current is the same at every point regardless of the order of components, and each component's voltage drop depends only on its own resistance (V = IR), not its position. Since the total resistance of the circuit is unchanged by reordering, the current stays the same, so the bulb glows with the same brightness. This shows that "series" means components share the same current path, and position in the loop doesn't affect the physics, only what's in the loop does.