Calculate electrical power (P = VI, P = I²R): Question bank

Before you start, keep these three plain-word anchors in mind, because every trap plays with one of them:

Power $P$ = energy per second (watts, $1 W = 1 J/s$ ).
Voltage $V$ = energy handed to each unit of charge (see Voltage and Potential Difference).
Current $I$ = amount of charge flowing per second (see Electric Current).
The trio $P = V I = I^{2} R = V^{2} / R$ are the same law wearing three costumes, stitched together by Ohm's Law (V = IR).

True or false — justify

True or false:

P = I^{2} R

and

P = V^{2} / R

are two different laws of nature.

False. Both come from

P = V I

after substituting Ohm's law

V = I R

; they are one law rewritten to fit the quantities you happen to know.

True or false: doubling the current through a fixed resistor doubles the heat it dissipates.

False. With

R

fixed,

V

also doubles (Ohm), so

P = I^{2} R

quadruples — two things doubled means ×4 power.

True or false: for a fixed resistor, halving the voltage across it cuts the power to one quarter.

True.

P = V^{2} / R

with

R

fixed, so power scales with voltage squared; half the voltage gives

(1/2)^{2} = 1/4

the power.

True or false: if two devices draw the same current, the one at higher voltage delivers more power.

True.

P = V I

; with

I

equal, larger

V

means more energy per coulomb, so more watts.

True or false: a resistor with a higher resistance always dissipates more power.

False. It depends on what's held constant. At fixed current

P = I^{2} R

rises with

R

; but at fixed voltage

P = V^{2} / R

falls as

R

rises. You must state what's fixed.

True or false: increasing

R

increases power in every case.

False. Same trap as above from the other side —

V^{2} / R

decreases with

R

. The relationship flips depending on whether current or voltage is the fixed quantity.

True or false: the watt already contains "per second," so multiplying watts by seconds gives joules.

True.

1 W = 1 J/s

, so

E = P t

recovers joules (see Energy and the Joule) — but only if

t

is in seconds.

True or false: a 100 W bulb and a 100 W heater deliver the same rate of energy.

True. A watt is a watt regardless of the device — both convert 100 joules of electrical energy every second, though into different useful outputs.

Spot the error

"Find the heat in a

100 Ω

resistor carrying

0.05 A

: since

P = V^{2} / R

, and the supply is

12 V

P = 144/100 = 1.44 W

The error is using the supply voltage instead of the voltage across that resistor. With only

I

and

R

known, use

P = I^{2} R = (0.05)^{2} \times 100 = 0.25 W

"A device runs for 2 hours at 60 W, so energy

= 60 \times 2 = 120 J

Time must be in seconds for joules.

E = 60 \times (2 \times 3600) = 432, 000 J

; the "2" was hours, not seconds.

"To reduce heat loss in a long cable, use a thinner wire so less current fits through."

Backwards. Thinner wire means higher resistance, and

P = I^{2} R

heat rises with

R

for the same current. You want thick, low-

R

wire (see Heat Dissipation and Cooling).

"Since

P = V I

is linear in

I

, tripling the current triples the resistor's heat."

In a fixed resistor,

V

is not independent — it rises with

I

. Use

P = I^{2} R

: tripling current gives

3^{2} = 9 \times

the power, not 3×.

"A 0.25 W resistor is safe as long as the voltage across it never exceeds 0.25 V."

Rating is about power, not voltage. Safety depends on

V^{2} / R

I^{2} R

; a small voltage across a small resistance can still exceed 0.25 W.

"Two resistors in series share the supply voltage, so I can put the full supply voltage into

V^{2} / R

for each one."

In series the voltage divides across the resistors, so each sees only its own share. Using the full supply voltage overstates each resistor's power.

P = V I

and

P = I^{2} R

must give different answers because one has a square and one doesn't."

They give the same answer for the same situation;

I^{2} R

is just

V I

after substituting

V = I R

. If they disagree, a wrong value was plugged in.

Why questions

Why does the

t

(time) vanish when we derive

P = V I

from

P = E / t

Because

E = V Q = V (I t)

, so

P = V (I t) / t

and the

t

cancels — energy-per-second collapses cleanly into voltage times current.

Why does

P = I^{2} R

have a square on the current but

P = V I

does not?

I^{2} R

the resistor is fixed, so raising current also raises the voltage drop (

V = I R

). Both factors of

V I

grow with

I

, giving

I \times I = I^{2}

Why do power companies transmit electricity at very high voltage?

For a given amount of delivered power, high voltage means low current, and heat loss in the lines is

I^{2} R

— cutting current dramatically cuts wasted heat.

Why does a resistor's power rating matter more than just its resistance value?

The rating tells you how much heat it can shed before failing; exceed the

I^{2} R

(or

V^{2} / R

) it can dissipate and it overheats and burns (see Resistors and Power Ratings).

Why can we say all three formulas are "the same law"?

Each is

P = V I

with one variable swapped using Ohm's law

V = I R

. They describe identical physics; only which two quantities you know changes which costume is convenient.

Why does choosing the right formula depend on what you already know rather than what's "correct"?

All three are correct, but each avoids extra work: with

V, I

use

V I

; with

I, R

use

I^{2} R

; with

V, R

use

V^{2} / R

— so you never have to compute a missing quantity first.

Edge cases

What is the power when the current is exactly zero (an open circuit)?

Zero watts.

P = V I = V \times 0 = 0

; no charge flows, so no energy is delivered no matter how high the voltage.

What is the power dissipated when the voltage across a resistor is exactly zero (both ends at the same potential)?

Zero watts.

P = V^{2} / R = 0

; with no potential difference there's no push, so no charge does work and no heat appears.

What does

P = V^{2} / R

predict as

R

shrinks toward zero (a short circuit), and is it physical?

It predicts

P \to \infty

. In reality tiny stray resistances and the source's own limits cap it, but the formula correctly warns that a short releases enormous power — hence sparks and blown fuses.

What does

P = I^{2} R

predict as

R \to 0

for a fixed current?

P \to 0

. A perfect conductor carrying current dissipates no heat, which is why superconducting or very-thick paths are prized for reducing losses.

If a device is rated exactly at its limit — say a 0.25 W resistor dissipating exactly 0.25 W — what's the risk?

It runs at the very edge with no thermal margin; small rises in current, voltage, or ambient temperature push it over, so good design leaves headroom.

Can power be delivered with a huge voltage but almost no current, or vice versa?

Yes —

P = V I

only cares about the product. High-

V

/low-

I

(transmission lines) and low-

V

/high-

I

(a starter motor) can give the same wattage.

Connections

Ohm's Law (V = IR) — the substitution that links all three power forms.
Voltage and Potential Difference — why zero voltage means zero power.
Electric Current — why zero current means zero power.
Energy and the Joule — the $E = P t$ trap with seconds vs hours.
Resistors and Power Ratings — the edge-of-limit and rating traps.
Heat Dissipation and Cooling — the $I^{2} R$ thick-wire and transmission logic.