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

CHAPTER 11
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).
– reciprocal of inductance.
Capacitance (C) – it is the ability to store electrical charge.
Its unit is Farad (F).
Elastance – a measure on how “susceptible” an element is to the
passage of current thru it. And it is the reciprocal of
capacitance.
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 = e x i
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
EmIm / 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 = Emsinwt
i = Im sin(wt - 2/‫)ח‬
L
b. Wave Diagram
i p
ω t
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 sin wt = L di/dt
Rearranging the equation becomes;

di = (Em/L) sin wt dt
By integration;
i = (Em/wL) sin(wt - 2/‫)ח‬

or;
i = Im sin(wt - 2/‫)ח‬
Where:
Im = Em/wL ; XL = 2‫ח‬fL = wL
The instantaneous power (p):
p = e x I = (Em sinwt) x (Im coswt)

p = - EI sin2wt
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 2/‫ ח‬to ‫ ח‬radians.
From the instantaneous power equation;

that is, p = - EI sin2wt
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
Example 3 p. 300]
When a 117-volt 60-cycle source is connected to a pure
inductor, the current is 3.9 amp. (a) Write the equation for the
current in the circuit. Determine (b) the energy stored in the
inductor bet wt1 = 2/‫ ח‬and wt2 = ‫( ;ח‬c) the stored energy bet wt1
= 4/‫ ח‬and wt2 = 34/‫( ;ח‬d) the energy stored in the magnetic field
bet wt1 = 4/‫ ח‬and wt2 = ‫ח‬.
The Capacitance (C) – Circuit
a. Circuit Diagram
e = Emsinwt
i = Im sin(wt + 2/‫)ח‬
b. Wave Diagram
e
p i
0 ‫ח‬2 ‫ח‬ 2/‫ח‬
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,
t
E = iR + ∫o idt/C
When applied to ac circuits, where R is zero and impressed emf

is a sinusoidal function, it becomes;
e = Em sin wt = 1/C ∫ot idt
Rearranging the equation becomes;
q = CEm sin wt
Getting the derivatives;
dq/dt = wCEm coswt

or;
i = wCEm sin (wt + 2/‫)ח‬

Where:
Im = wCEm ; XL = 2‫ח‬fL = wL
The instantaneous power (p):
p = e x i = (Em sinwt) x wCEm sin (wt + 2/‫)ח‬

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/‫ח‬
radians.
From the instantaneous power equation;

that is, p = EI sin2wt
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
Example 3 p. 300
The current in a circuit is 1.96 amp when a capacitor is
connected to a 250-volt 50-cycle source. (a) Write the current and
power equations. (b) Determine the energy stored in the capacitor
during the positive half of the power wave.
The Series Inductance-Capacitance (L-C) Circuit.
Circuit Diagram
E
I
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?

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CHAPTER 11

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

Resistance (R) – is the opposition of current flow in a conductor.

Basic Types of Circuit:

1. The Resistance (R) – circuit

The instantaneous power (p) in the resistance circuit:

By proper substitution, the equation may be rewritten in the ff.

2. The Inductance (L) – circuit

When a sinusoidal voltage is impressed across a pure

Based from Chapter 8, Current Growth in Inductive Circuits,

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

Rearranging the equation becomes;

i = (Em/wL) sin(wt - 2/‫)ח‬

The instantaneous power (p):

p = e x I = (Em sinwt) x (Im coswt)

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

From the instantaneous power equation;

It follows that the differential energy delivered to the

Substituting the instantaneous power p and by integrating

W = LI2 joules (Formula)

W = energy stored in the inductor, Joules

The Capacitance (C) – Circuit

0 ‫ח‬2 ‫ח‬ 2/‫ח‬

When a sinusoidal voltage is impressed across a pure

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

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

Rearranging the equation becomes;

Getting the derivatives;

dq/dt = wCEm coswt

i = wCEm sin (wt + 2/‫)ח‬

The instantaneous power (p):

p = e x i = (Em sinwt) x wCEm sin (wt + 2/‫)ח‬

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

From the instantaneous power equation;

It follows that the differential energy delivered to the

Substituting the instantaneous power p and by integrating

The Series Inductance-Capacitance (L-C) Circuit.

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

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

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