* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Circuit Sums with ac
Regenerative circuit wikipedia , lookup
Crystal radio wikipedia , lookup
Negative resistance wikipedia , lookup
Josephson voltage standard wikipedia , lookup
Spark-gap transmitter wikipedia , lookup
Radio transmitter design wikipedia , lookup
Schmitt trigger wikipedia , lookup
Index of electronics articles wikipedia , lookup
Operational amplifier wikipedia , lookup
Valve RF amplifier wikipedia , lookup
Electrical ballast wikipedia , lookup
Opto-isolator wikipedia , lookup
Current source wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
RLC circuit wikipedia , lookup
Surge protector wikipedia , lookup
Power electronics wikipedia , lookup
Current mirror wikipedia , lookup
Power MOSFET wikipedia , lookup
Circuit Sums with a.c. If the circuit only contains resistances (as well as generators of voltage), the calculations work just like d.c. Fortunately, in the domestic situation, such circuits are, near enough, the norm. Complications arise if the circuit includes Inductance and/or Capacitance, both of which cause the current and voltage not to be "in phase" with each other. Unfortunately, electric motors do display inductance ! If we have a circuit including inductance, we find that the current has its peaks and troughs a little later than those of the voltage; we describe this situation by saying that the current LAGS the voltage. If the circuit ONLY contains inductance, the lag is quarter of a cycle of the a.c., but it is less if the circuit contains a combination of resistance and inductance (as is more usual than pure inductance). 10 8 6 4 2 0 -2 -4 -6 -8 -10 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 The "x" axis is in seconds and the "y" axis is in volts (voltage) and ampere (current). 0.09 0.1 The largest trace (10 V p-p) is the voltage, whilst the other two are the currents we would expect in a purely inductive circuit (A) and in a circuit containing both inductance and resistance (B). What is the PHASE SHIFT or PHASE DIFFERENCE between the voltage and current in each case ? (1 full cycle = 360 degree). If the circuit contains capacitance, the opposite effect happens; the current actually happens earlier than the voltage, or LEADS it as the saying goes. We often make use of that fact in practice to reduce the phase lag of systems including inductive motors. I wonder why we might want to do so ? Read on ... The Power Factor With direct current and voltage, the power was just V x I. It is more complicated with a.c. because of the phase difference. It is V x I x cos() where is the phase difference (expressed as an angle) between the current and the voltage. The influence of the phase difference is that, at a given voltage, we need more current for the same power if there is a phase difference. The costs of providing the transformers and cables the electricity supply company needs to use (including the "I2R" power losses in the cables) are strongly dependent on the current, so we are unpopular if our power factor is low. Industrial electricity tariffs charge us per kwh (just like domestic ones) but also on installed capacity in kVA (kilovolt-ampere) and, in large installations, also on maximum demand in kVA. We can reduce the "A" if we use capacitors to raise the power factor so we have fewer amperes for the same power. The Impedance An a.c. quantity corresponding to resistance in d.c. circuits. For inductance, it is equal to 2fL where f is the frequency in hertz and L is the inductance in henry. For capacitance, it is equal to 1/(2fC) where f is the frequency in hertz and C is the capacitance in farads. If we have a series circuit containing more than one of the above, we have to add them up as if they are Vectors (we actually call them "Phasors"). Inductive reactance Resistance Capacitative reactance So if we want to calculate the current in a circuit consisting of a 240-V power supply, a 50-ohm resistance, and an 0.3-henry inductance, we do the following sum: Impedance = 50 ohm resistive + (0.3 x 2 x 50) ohm inductive Adding it up vectorially: 0.3 x 2 x 50 = 94.2 ohm X 50 ohm Send for Pythagoras ! 502 + 94.22 = 11374; the resultant is the square root of this quantity, i.e. 106.6 ohm. The current is therefore 240/106.6 = 2.25 A. We can also work out the power, as the angle X gives us the phase difference between the current and the voltage. The tangent of the angle X is side opposite/side adjacent = 94.2/50 = 1.884. The angle whose tangent is 1.884 (notation arctan(1.884) or tan -1(1.884)) is 62 degrees (1.083 radian). So the power is 240 x 2.25 x cos(62 degrees) = 253.2 watt. We could also have calculated the power by noting that no power is consumed in a pure inductance (why ?) so all the power consumed will be in the resistance. Using I2R, we obtain 2.252 x 50 = 253.1 watt (the missing 0.1 watt is explained by numerical rounding errors in the calculation). Some Questions for You What will the circuit current and power be if : a) We connected a 30-uF capacitor instead of the inductor. b) We connected a 30-uF capacitor in series with the inductor. c) We connected just an inductor of 0.5 H with a resistance of 10 ohm.