Download Circuit Sums with ac

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 2fL where f is the frequency in hertz and L is the
inductance in henry.
For capacitance, it is equal to 1/(2fC) 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.