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Circuit energy cost: current to power to kilowatt-hours

SPH3U Grade 11 Physics · Electricity and Magnetism

A student stripping paint in an unheated garage sets a portable shop heater beside the workbench and plugs it into the 120 V outlet. Behind the grille sit two nichrome coils, marked 15.0 Ω and 9.0 Ω on the nameplate, and the wiring diagram on the back panel shows that the LOW setting connects the two coils in series across the supply. The student flips the selector to LOW, watches the coils come up to a dull orange glow, and goes back to work. The heater runs untouched for 6.00 h before it is finally switched off. Find the total resistance of the series pair, the current the heater draws, the power it dissipates, and the electrical energy it consumes over the run.

3.60 kWh5.00 A120 V15.0 Ω9.0 Ω600 W on LOW for 6.00 h

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • V = 120 V — Outlet voltage
  • R₁ = 15 Ω — First nichrome coil
  • R₂ = 9 Ω — Second nichrome coil
  • t = 6 h — Time left running on LOW
Determine
  1. (a)the total resistance of the series pair
  2. (b)the current the heater draws
  3. (c)the power it dissipates
  4. (d)the electrical energy it consumes over the run
Step 1 of 4(a) · solve for Total resistance

On LOW the coils sit end to end, so their resistances add. Nothing else about the circuit can be computed until this total is known.

R1R2Rt
Rearranged for R_t
Rt=R1+R2R_{t} = R_{1} + R_{2}
Your values, in your units
Rt=(15 Ω)+(9 Ω)R_{t} = \left( 15\ \Omega \right) + \left( 9\ \Omega \right)
Answer
Rt=24 ΩR_{t} = 24\ \Omega

Carried onward at full precision, not this rounded figure.

Open the Two Resistors in Series solver →

Step 2 of 4(b) · solve for Current

The outlet holds 120 V across the pair. Ohm's law turns that into the single current every part of a series loop shares.

IRV
Rearranged for I
I=VRI = \tfrac{V}{R}
24 Ωcarried from step 1
Your values, in your units
I=(120 V)(24 Ω)I = \tfrac{\left( 120\ \text{V} \right)}{\left( 24\ \Omega \right)}
Answer
I=5 AI = 5\ \text{A}

Carried onward at full precision, not this rounded figure.

Open the Ohm's Law solver →

Step 3 of 4(c) · solve for Power

Power from the current and the resistance — the first step that draws on two earlier answers at once.

IRP
Rearranged for P
P=I2RP = I^{2} R
5 Acarried from step 2
24 Ωcarried from step 1
Your values, in your units
P=(5 A)2 (24 Ω)P = \left( 5\ \text{A} \right)^{2} \, \left( 24\ \Omega \right)
Answer
P=600 WP = 600\ \text{W}

Carried onward at full precision, not this rounded figure.

Open the Electrical Power (P = I²R) solver →

Step 4 of 4(d) · solve for Energy

Power sustained over time is energy. The answer lands in joules; the utility meters the same quantity in kilowatt-hours.

EPt
Rearranged for E
E=PtE = P t
600 Wcarried from step 3
Your values, in your units
E=(600 W) (6 h)E = \left( 600\ \text{W} \right) \, \left( 6\ \text{h} \right)
Converted to base units
E=(600 W) (21,600 s)E = \left( 600\ \text{W} \right) \, \left( 21{,}600\ \text{s} \right)
Answer
E=12.96 MJE = 12.96\ \text{MJ}

Carried onward at full precision, not this rounded figure.

Open the Electrical Energy (E = Pt) solver →

Answer

Therefore the series pair totals 24.0 Ω, the heater draws 5.00 A and dissipates 600 W on LOW, and six hours of that consumes 12.96 MJ — the 3.60 kWh the utility meter would record.

Why this order

Every electricity bill is this chain. The supply voltage is fixed by the utility and the resistance is fixed by the appliance, so current is the first thing you can actually compute — nothing downstream is knowable until Ohm's law hands it over. Step 3 is where the chain stops being a straight line: P = I²R reaches back for the current from step 2 and the same total resistance from step 1. P = VI would have given the identical 600 W, and picking between them is really a question of which two quantities you measured rather than inferred.

The trap is the series setting. Two coils end to end share one current and their resistances add to 24 Ω — more opposition than either coil alone — so the heater draws less power on LOW, not more, which is exactly the point of the switch. Students routinely reason that two elements must mean more heat. The second trap is units: joules are the SI answer, but nobody bills in them. 12.96 MJ is 3.60 kWh, and at 12.5 cents a kilowatt-hour that garage cost about 45 cents to warm — the kind of arithmetic that turns an efficiency argument from a slogan into a number.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.