Resistance (R), Inductance (L), and Capacitance (C) Circuits

Definition of Terms:

Resistance (R) – is the opposition of current flow in a conductor. Its unit is ohms (Ω).
- It is the ratio of voltage to current for constant voltage and current.
Conductance (G)– a measure of how well the material will conduct electricity. It is the
reciprocal of resistance.
Inductance (L) – is a property, which resists changes in current. Its unit is Henry (H).
Capacitance (C) – it is the ability to store electrical charge. Its unit is Farad (F).
Susceptance (B) – a measure on how “susceptible” an element is to the passage of
current thru it. And it is the reciprocal of capacitance and inductance.
Inductive Reactance (XL) – is the opposition to current flow, which results in the continual
interchange of energy between the source and the magnetic field of the inductor.
- it is also the property of an inductor that makes the current to lag
behind the voltage by 90 elec deg. (For purely inductive load only)
Capacitive Reactance (XC) – is the opposition to the flow of charge, which results to the
continual interchange of energy between the source and the electric field of the
capacitor.
- it is the property of capacitor that makes the current to lead the
voltage by 90 elec deg. (For purely capacitive circuit only.)
Impedance (Z) – is the geometric sum of the IR and IXL voltage drops (for R-L circuit)
and IR and IXC drops (for R-C circuits.
- it is measured in Siemens (S) or Mho.
Admittance (Y) – a measure of how easily a network will “admit” the passage of current
thru the system. And it is the reciprocal of impedance.
Impedance diagram – a vector display that clearly depicts the magnitude of the
impedance of the resistive, reactive, and capacitive components of a network,
and the magnitude and angle of the total impedance of the system.
Phasor – a radius vector that has a constant magnitude at a fixed angle from the positive
real axis and that represents a sinusoidal voltage or current in the vector domain.
Phasor Diagram – a vector display that provides at a glance the magnitude and phase
relationships among the various voltages and current of a network.
Basic Types of Circuit:
The study of circuits involves three basic types of units. These are the following:

1. The Resistance (R) – circuit

a. Circuit Diagram:
e = Emsinwt
i = Imsinwt

R
b. Wave Diagrams

0 Π/2 3Π/2 2Π

c. Phasor Diagram
I E

The instantaneous power (p) in the resistance circuit:

The instantaneous power (p) in the purely resistive circuit can be expressed by the
formula…
p=exi
Substituting e and i;
p = (Emsinwt) x (Imsinwt) = EmImsin2wt
But: Sin2wt = ½ - ½ cos2wt
By proper substitution, the equation may be rewritten in the ff. form:
p = EmIm ( ½ - ½ cos2wt)
= (EmIm/2) – [(EmIm/2)cos2wt]
Note:
“The average power per cycle represented by the cos2wt term is zero because the
positive and negative halves of each cycle of a cosine (or sine) function cancel each other.
This means therefore that the constant term, represented by E mIm / 2, is the
average power delivered to the resistor.

That is;
Pave = EmIm / 2 = EI watts (Formula 1)

Example 01/297
An incandescent lamp load, generally considered to be made up of resistors, takes
4.8 kw from a 120-volt ac source. Calculate (a) the total current, (b) the instantaneous
value of power, (c) the resistance of the load.

2. The Inductance (L) – circuit

a. Circuit Diagram

e = Emsinωt

i = Im sin(ωt - π/2)

b. Wave Diagram

i p

ω
 c. Phasor Diagram

90o E

When a sinusoidal voltage is impressed across a pure inductance, the current wave
will also be sinusoidal. However, unlike the pure resistance circuit in which e and i are
in phase, the current will lag behind the voltage by 90 elec deg.

Based from Chapter 8, Current Growth in Inductive Circuits, it was shown that a
direct current builds up in an R-L circuit in accordance with the equation;

E = iR + L di/dt

When applied to ac circuits, where R is zero and impressed emf is a sinusoidal

function, it becomes;

e = Em sinωt = L di/dt

Rearranging the equation becomes;

di = (Em/L) sinωt dt

By integration;

i = (Em/ωL) sin(ωt - π/2)

or;

i = Im sin(ωt - π/2)

Where:

Im = Em/ωL ; XL = 2πfL = ωL
The instantaneous power (p):

p = e x I = (Em sinωt) x (Im cosωt)

p = - EI sin2ωt

Where:
Pave = 0 for a purely inductive circuit.
Pmax = EI

Example 2 p.299]
An inductance of 0.106 Henry is connected to a 120-volt 60-cycle source. Calculate
(a) the inductive reactance, (b) the current in the circuit, (c) the average power taken by
the inductor, (d) the maximum power delivered to the inductor or returned to the source.
Write the equations for (e) the current, and (f) the power.

Energy in an Inductive Circuit:

To determine the energy in joules(watt-sec) that is stored in the inductor, it is

important to understand that this storage takes place in continually changing increments
of p dt as the current rises from zero to a maximum. This is between π/2 to π radians.

From the instantaneous power equation;

that is, p = - EI sin2ωt

It follows that the differential energy delivered to the inductor in time dt is…

dW = p dt

Substituting the instantaneous power p and by integrating this equation;

W = LI2 joules (Formula)

Where:

W = energy stored in the inductor, Joules

L = inductance of the circuit, Henry
I = current in Amp

The Capacitance (C) – Circuit

a. Circuit Diagram

e = Emsinωt

i = Im sin(ωt + π/2)

b. Wave Diagram

e
p i

0 π/2 π 2π
 c. Phasor Diagram

90o E

When a sinusoidal voltage is impressed across a pure capacitance, the current

wave will also be sinusoidal. However, unlike the pure resistance circuit in which e
and i are in phase, the current will lead the voltage by 90 elec deg.

Based from Chapter 9, Charging Current in an RC Circuits, and in accordance with

Kirchhoff’s Law,

E = iR + ∫o t idt/C

When applied to ac circuits, where R is zero and impressed emf is a sinusoidal

function, it becomes;

e = Em sin ωt = 1/C ∫ot idt

Rearranging the equation becomes;

q = CEm sin ωt

Getting the derivatives;

dq/dt = wCEm cosωt

or;

i = ωCEm sin (ωt + π/2)

Where:

Im = ωCEm ; Xc = 1/ωC = 1/2πfC

The instantaneous power (p):

p = e x i = (Em sinωt) x ωCEm sin (ωt + π/2)

p = EI sin2wt

Where:
Pave = 0 for a purely capacitive circuit.
Pmax = EI
Example 4 p.304]
A 127-μf capacitor is connected to a 125-volt 50-cycle source. Calculate (a) the
capacitive reactance, (b) the current in the circuit, (c) the average power taken by the
capacitor, (d) the maximum power delivered to the capacitor or returned to the source.
Write the equations for (e) the current and (f) the power.

Energy in a capacitive Circuit:

To determine the energy in joules (watt-sec) that is stored in the capacitor, it is

important to understand that this storage takes place in continually changing increments
of p dt as the current rises from zero to a maximum. This is between 0 to 2/‫ ח‬radians.

From the instantaneous power equation;

that is, p = EI sin2ωt

It follows that the differential energy delivered to the capacitor in time dt is…

dW = p dt

Substituting the instantaneous power p and by integrating this equation;

W = CE2 joules (Formula)
Where:
W = energy stored in the capacitor, Joules
C = capacitance of the circuit, Henry
I = current in Amp
The Series Inductance-Capacitance (L-C) Circuit.

Circuit Diagram
E

L C

Formula:

If XL > XC (The series L-C behave like an inductance)

a. Xeq = XL - XC

If XC > XL (The series L-C behave like a capacitor)

b. a. Xeq = XC - XL
Note: “an increase in L or C will result in an over-all increase in the inductive reactance
of such a circuit; a decrease in L or C will, on the other hand, result in an over-all increase
in the capacitive reactance of a similar circuit.”

Ex. 06 p.306
A series circuit consisting of a 0.0795-henry inductor and a 177-F capacitor is
connected to a 120-volt 60-cycle source. Calculate (a) the equivalent reactance of the
circuit, (b) the circuit current, indicating whether the latter lags or leads.

Ex. 08 p. 306
 A series circuit consisting of a 0.0795-henry inductor and a 177-F capacitor is
connected to a 120-volt variable frequency source. At what frequency will the circuit take
a lagging current of 4 amp?

The Series Resistance-Inductance (R-L) Circuit.

Circuit diagram
E

R L
Phasor Diagram

EL = IXL E =  ER2 + EL2

900
θ
ER = IR I
Working Formulas:

E=IxZ ; Z =  R2 + XL2
Cos  = p.f. = ER / E = IR / IZ = R/Z = P / EI
e = Em sinα ; i = Im sin (α-)
P = EI cos 
Ex. 12 p. 311
A load of 18.4-kw operating at a power factor of 0.8 lagging is connected to a 460-
volt 60-cycle source. Calculate (a) the load current; (b) the power-factor angle; (c) the
equivalent impedance, resistance, and reactance of the load. (d) Write the equations for
the voltage and current.
Prob. 19 p. 322
An impedance coil has a resistance of 7.5-ohms and an inductive reactance of 18-
ohms. (a) What is the equation of the voltage wave that produces a current i = 11.3 sin
wt? (b) Calculate the values of of E, I, ER and EL.

The Series Resistance-Capacitance (R-C) Circuit.

a. Circuit Diagram
E

R C

b. Phasor Diagram
ER = IR I
θ

Ec = IXc E =  ER2 + Ec2

Working Formulas:
E=IxZ ; Z =  R2 + XC2
Cos  = p.f. = ER / E = IR / IZ = R/Z = P / EI
e = EM sinα ; i = IM sin (α+)
P = EI cos 
Ex. 13 p. 312
A 125-volt 25-cycle source is connected to a series circuit consisting of a 30-ohm
resistor and a 159-µF capacitor. Calculate the ff: impedance, current, power factor,
power.

Prob. 29 p. 322
A series R-C circuit takes a current whose equation is I = 0.85 sin (754t + п/4)
when connected to a source of emf having the equation e = 340 sin 754t. Calculate (a)
the values of Z, R and XC,; (b) the capacitance of the capacitor; (c) the circuit power
factor and power.

The Series Resistance-Inductance-Capacitance (R-L-C)

Circuit
a. Circuit Diagram
E

R L C
b. Phasor Diagram

At XL > XC
XL Z =  R2 + (XL – XC)2

900
θ
R I
At XC > XL
R I
θ

Xc Z =  R2 + (Xc – XL)2
Working Formulas:
At XL > XC ; The circuit acts like an R-L circuit and the current is lagging.
Z = √ R2 + (XL – XC)2
At XL < XC ; The circuit acts like an R-C circuit and the current is leading.

Z = √ R2 + (XC – XL)2
E=IxZ ; P = EI cos 
Note:
At XL = XC, the circuit is said to produce resonance. And the circuit is like a purely
resistive circuit.

That is;
Z = R and XL = X C

Ex. 15 p. 316
A series circuit consisting of an 80-ohm resistor, a 0.3-henry inductor, and a 50-
µF capacitor is connected to a 120-volt 60-cycle source. Calculate the ff: (a) equivalent
impedance of the circuit, (b) current, (c) voltage drops across the several units, (d) power
and power factor.
Volt-amperes and Reactive Volt-amperes

The Power Triangle Diagram:

a. Inductive load
b. Capacitive load
Power Formulas:
a. True, Real or Active Power (P)
P = EI cos  = I2R watts
b. Reactive or Idle Power (Q)
Q = EI sin  = I2X vars or reactive volt-amperes
Reactive Factor (RF) = sin  = Q / S
c. Apparent Power (S)
S = EI = I2Z Volt-amperes
Ex. 18 p. 319
A load of 250 KVA, operating at a power factor of 0.86 lagging, is connected to a
2,300-volt a-c source. Calculate (a) power, (b) current, (c) reactive kilovolt-amperes, (d)
reactive factor.

Supplementary Problems:

Prob. 37 p. 323
A 46-kv 0.8-lagging power factor load is connected to the end of a short
transmission line where the voltage is 230. If the line resistance and reactance are 0.06
and 0.08 ohm, respectively, calculate the voltage at the sending end.

Prob. 41 p. 323
A telephone receiver has an impedance of 306 ohms at 800 cps and a resistance
of 60 ohms. For what value of capacitance of a capacitor in series with the coil will the
power factor be unity at 1,000 cps?

B.P. Oct ‘98

Two relays each with 20-ohm resistance and 0.16-henry inductance are connected
in series. What is the equivalent impedance?
a. 20 + j102.2 ohms c. 20 + j95.32 ohms
b. 20 + j120.63 ohms d. 10 + j25.32 ohms

B.P. Mar. ‘98

A 50-µF and 100-µF capacitors are connected in series and across a 100 sin (wt +
30 ) voltage. Write the equation of the current.
o

a. 1.26 sin (wt + 120o) c. 5.65 sin (wt + 120o)

b. 1.26 sin (wt + 90o) d. 5.65 sin (wt + 90o)s

Math B.P. Apr. ‘97

If e = 100sin(wt+30o) – 50 cos3wt + 25 sin(5wt+150o) and i = 20 sin(wt+40o) +
10 sin(3wt + 30o) – 5 sin(5wt-50o). Calculate the power in watts.
a. 1177 c. 1043
b. 937.6 d. 1224