Ohm's law at power
Anumang dalawa sa V, I, R at P ay magbibigay ng iba pang dalawa.
Subukan ang isang job 4
Nilalagyan ng halimbawang value ang bawat job. Palitan ang mga ito ng sarili mong value.
- Find the resistance of a heater elementThe working resistance and current of a 2 kW heater on 230 V.
- Find the current of a 12 V lampThe current a 50 W lamp draws on 12 V d.c.
- Find the power of a three‑phase heater bankThe kW of a star heater bank of 20 Ω per phase on 400 V.
- Find the voltage across a resistorThe volts across 470 Ω carrying 20 mA d.c.
Mga input
Worked example gamit ang mga default na input.
- Supply
- Single‑phase a.c.
- Alam mo na
- Boltahe at power
- Voltage
- 230 V
- Real power
- 2,300 W
- Power factor
- 1
Current
10A
- Impedance
- 23Ω
- Apparent power
- 2.3kVA
Design aid lang ang resultang ito. I-verify ito laban sa kasalukuyang standard.
Mga hakbang ng kalkulasyon
- Formula
I = P ÷ (V × pf)
- Voltage230V
- CurrentResulta10A
- Impedance23Ω
- Real power2,300W
- Apparent power2.3kVA
- Reactive power0kvar
- Resistance part, Z × pf23Ω
- Reactance part, Z × sin φ0Ω
Tungkol sa calculator na ito
How this is calculated
Ohm's law says the voltage across a resistance is the current through it times the resistance. The power formula says the power is the voltage times the current. Together they link four quantities, so any two of them give the other two. You pick the pair you know, and the tool solves for the rest.
On d.c. the four are voltage, current, resistance and power, and nothing else enters. On a.c. the current and voltage can be out of step, so two things change. The opposition to current is the impedance, Z, which the tool uses wherever d.c. uses R. The real power is the voltage times the current times the power factor. The tool then splits the impedance into its resistance part, Z times the power factor, and its reactance part, and gives the apparent and reactive power too.
On three phase the tool takes the line voltage, between two phases, and the line current. The real power is the square root of three times the line voltage times the line current times the power factor. The impedance it gives is one phase of a balanced star load: the voltage from line to neutral over the line current. A delta load with the same line current has three times that impedance in each branch.
No standard is read. These are the physics every electrician learns, and the tool shows the formula it used for the pair you gave.
| Step | Table or clause | What it decides |
|---|---|---|
| d.c. | V = I × R and P = V × I | The two unknowns from the pair you know |
| Single‑phase a.c. | V = I × Z and P = V × I × pf | The current or voltage, the impedance and the real power |
| Three‑phase a.c. | V = √3 × I × Z and P = √3 × V × I × pf | Line values and the impedance per phase of a star load |
| Power from resistance and current | P = I² × R | The power when the voltage is not given |
| The parts of the impedance | R = Z × pf and X = Z × sin φ | The resistance and reactance on a.c. |
Worked examples
- A 2 kW heater on 230 V draws 8.7 A. Its working resistance is 26.45 Ω.
- A 50 W lamp on 12 V d.c. draws 4.17 A through 2.88 Ω.
- A star heater bank of 20 Ω per phase on 400 V draws 11.55 A in each line and gives 8,000 W.
- A 470 Ω resistor carrying 20 mA has 9.4 V across it and gives off 0.188 W.
- A motor on 400 V drawing 32 A at 0.85 power factor looks like 7.22 Ω per phase and takes 18,844.71 W.
Limits
The tool treats every load as linear and balanced. It does not model harmonics, an unbalanced three‑phase load or a load whose resistance changes with temperature, such as a lamp filament or a cold heater element. A meter reading on a cold element is lower than its working value.
For a cable's voltage drop, use the voltage drop tool: it reads the cable's resistance and reactance at its operating temperature from AS/NZS 3008.1.
See also: Three phase power, Full load current and Voltage drop.
Mga tanong
Design aid ang resulta. I-verify ito laban sa kasalukuyang standard.
Ano ang Ohm's law?
Ang voltage sa ibabaw ng resistance ay katumbas ng current na dumadaan dito na i-multiply sa resistance: V = I × R. Sa P = V × I, nauugnay nito ang lahat ng apat na quantity, kaya ang alinmang dalawa ay nagbibigay ng iba pang dalawa.
Paano binabago ng power factor ang resulta sa a.c.?
Sa a.c., ang hadlang sa current ay ang impedance Z, hindi lang ang resistance. Ang real power ay V × I × power factor sa isang phase. Ibinibigay din ng tool ang resistance part, Z × pf, at ang reactance part.
Paano gumagana ang tool sa three phase?
Kinukuha nito ang line voltage at line current. Ang real power ay √3 × V × I × power factor. Ang impedance na ibinibigay nito ay isang phase ng balanced star load. Ang delta load ay tatlong beses nito sa bawat branch.
Bakit mas mababa ang resistance na nababasa ng heater element kapag malamig?
Tumataas ang resistance ng metal element habang tumataas ang temperatura nito. Mas mababa sa working resistance ang cold reading sa meter. Gamitin ang rated volts at watts para sa working value.
Puwede ko bang gamitin ito para sa voltage drop ng cable?
Tantiya lang. Nagbabago ang resistance ng cable ayon sa temperatura, at nagdaragdag ng reactance ang a.c. Gamitin ang voltage drop tool, na bumabasa ng AS/NZS 3008.1 para sa cable.
Mga worked example
Background
Mga kaugnay na tool
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