Current and power conversion formula - Solutions - Huaqiang Electronic Network

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Current I, voltage V, resistance R, power W, frequency F
W=I squared multiplied by R
V=IR
W = the square of V divided by R

Current = voltage / resistance power = voltage * current * time

Current I, voltage V, resistance R, power W, frequency F
W=I squared multiplied by R
V=IR

Current I, voltage V, resistance R, power W, frequency F
W=I squared multiplied by R
V=IR
W = the square of V divided by R

The relationship between voltage V (volts), resistance R (ohms), current intensity I (amperes), and power N (watts) is:
V=IR, N=IV = I*I*R,
Or it can be transformed into: I=V/R, I=N/V, etc. But it must be noted that all of the above are derived from the DC (more precisely, DC steady state) circuit! Applicable. If AC circuit, it should be supplemented and corrected to calculate the conversion relationship between voltage, resistance, current and power.

Current = I, voltage = U, resistance = R, power = P
U=IR, I=U/R, R=U/I,
P=UI, I=P/U, U=P/I
P=U2/R, R=U2/P
I remember this. I don't know if there is any P=I2R P=IU R=U/I. It is best to use these two; for example, the electric energy of the motor is converted into thermal energy and mechanical energy. Current symbol: I
Symbol Name: Ampere (An)
Unit: A
Formula: Current = Voltage / Resistance I = U / R
Unit conversion: 1MA (megaamperes) = 1000kA (kilograms) = 1000000A (an)
1A (A) = 1000mA (mA) = 1000000μA (microamperes) single-phase resistance type electric power calculation formula = voltage U * current I
Calculation formula of electric power of single-phase motor = voltage U* current I* power factor COSΦ
Calculation formula of three-phase resistance type electric power = 1.732* line voltage U* line current I (star connection method)
= 3* phase voltage U* phase current I (angle connection)
Calculation formula of three-phase motor electric power = 1.732* line voltage U* line current I* power factor COSΦ (star current = I, voltage = U, resistance = R, power = P
U=IR, I=U/R, R=U/I,
P=UI, I=P/U, U=P/I
P=U2/R, R=U2/P
I remember this, I wonder if there is still P=I2R (1) series circuit P (electric power) U (voltage) I (current) W (electric work) R (resistance) T (time)
The current is equal everywhere I1=I2=I
The total voltage is equal to the sum of the voltages across the consumers U=U1+U2
The total resistance is equal to the sum of the resistors R = R1 + R2
U1: U2=R1: R2
The total electric work is equal to the sum of each electric work W=W1+W2
W1: W2=R1: R2=U1: U2
P1: P2 = R1: R2 = U1: U2
The total power is equal to the sum of the powers P=P1+P2
(2) The total current of the parallel circuit is equal to the sum of the currents I=I1+I2
The voltage is equal everywhere U1=U1=U
The total resistance is equal to the product of each resistor divided by the sum of the resistors R = R1R2 ÷ (R1 + R2)
The total electric work is equal to the sum of each electric work W=W1+W2
I1: I2=R2: R1
W1: W2=I1: I2=R2: R1
P1: P2 = R2: R1 = I1: I2
The total power is equal to the sum of the powers P=P1+P2
(3) The electric power of the same electrical appliance 1 rated power than the actual power is equal to the rated voltage than the square of the actual voltage Pe/Ps = (Ue / Us) squared

2. Formula for circuit (1) Resistance R
1 resistance equals material density multiplied by (length divided by cross-sectional area) R = density × (L ÷ S)
2 resistance equals voltage divided by current R=U÷I
3 resistance is equal to the square of the voltage divided by the electric power R=UU÷P
(2) Electric work W
Electric work is equal to current multiplied by voltage multiplied by time W=UIT (general formula)
Electric work equals electric power multiplied by time W=PT
Electrical work is equal to charge multiplying voltage W=QT
Electric work is equal to the current squared multiplied by the time W = I × IRT (pure resistance circuit)
The electric work is equal to the square of the voltage divided by the resistance and multiplied by the time W=U•U÷R×T (ibid.)
(3) Electric power P
1 electric power is equal to voltage multiplied by current P=UI
2 electric power is equal to the square of the current multiplied by the resistance P = IIR (pure resistance circuit)
3 electric power is equal to the square of the voltage divided by the resistance P = UU ÷ R (ibid.)
4 electric power is equal to electric work divided by time P=W:T
(4) Electric heating Q
The electric heat is equal to the current squared into the resistance multiplied by the time Q=IIRt (the general formula)
The electric heat is equal to the current multiplied by the voltage multiplied by the time Q=UIT=W (pure resistance circuit power = 1.732 * rated voltage * current is the pure resistive load power calculation formula of the star connection in the three-phase circuit power = rated voltage * current is single Calculation formula of pure resistive load power in phase circuit
P=1.732×(380×I×COSΦ) is the inductive load power calculation formula of the star connection in the three-phase circuit.

Calculation formula of single-phase resistance electric power = voltage U* current I
Calculation formula of electric power of single-phase motor = voltage U* current I* power factor COSΦ
Calculation formula of three-phase resistance type electric power = 1.732* line voltage U* line current I (star connection method)
= 3* phase voltage U* phase current I (angle connection)
Calculation formula of three-phase motor electric power = 1.732* line voltage U* line current I* power factor COSΦ (star connection method)
= 3* phase voltage U* phase current I* power factor COSΦ (angular connection)
In the three-phase AC circuit, the two power calculation formulas of the star connection and the angle connection can be used interchangeably, but the phase voltage, the line voltage, the phase current, and the line current must be distinguished. Electric power calculation formula:
In pure DC circuit: P=UI P=I2R P=U2/R where: P---electric power (W), U---voltage (V),
I----current (A), R---resistance (Ω).
In single-phase AC circuit: P=UIcosφ where: cosφ---power factor, such as incandescent lamp, electric furnace, electric iron, etc. can be regarded as resistive load, its cos φ=1 then P=UI
U, I--- are phase voltage and current, respectively.
In a symmetrical three-phase AC circuit, regardless of the form of the load connection, the average power of the symmetrical three-phase load is:
P=√3UIcosφ where: U, I--- are line voltage and line current, respectively.
Cosφ --- power factor, if it is a three-phase resistive load, such as three-phase electric furnace, cosφ=1
Then P = √ 3UI.

W = the square of V divided by R