DC Networks Theorems

The laws that determine the currents and voltage drops in DC networks:

• Ohm’s law
• The law’s for resistors in series and parallel networks
• Kirchhoff’s laws

In addition to the above laws, we have several circuit theorems used to solve problems in electrical circuits and they include:

• The Superposition theorem
• Thevenin’s theorem
• Norton’s theorem
• The maximum power transfer theorem

Kirchhoff’s Laws

Kirchhoff’s laws states that :

Current law – At any junction in an electric circuit, the total current flowing towards that junction is equal to the total current flowing away from the junction i.e. ∑ I = 0

From figure above: I₁ + I₂ = I₃ + I₄+ I₅ or I₁ + I₂ -I₃ -I₄-I₅ = 0

Voltage Law –In any closed loop in a network, the algebraic sum of voltage drops i.e. products of current and resistance taken around the loop is equal to the resultant emf acting in that loop.

Let’s consider the diagram below:

E₁ + E₂ = IR₁ +IR₂ +IR₃

Note, if current flows away from positive terminal of a source, that source is considered to be positive. Hence, moving anticlockwise around the loop, E₁ is positive and E₂ is negative.

The Superposition Theorem

Superposition Theorem states that:

In a strictly linear network i.e. made up of linear resistances and containing more than one source of emf, the resultant current flowing in any branch is the algebraic sum of the currents that would flow in that branch if each source was considered separately, all other sources being replaced at the time by their respective internal resistances.

This principle applies to voltages and currents, but not powers which are current-voltage products.

Worked Example

To help us understand how this theorem is used in electrical circuits, let’s consider the circuit below containing two sources of emf each with internal resistance:

Solution

• Redraw the original circuit with source E₂ removed, and replaced by r₂ only.

• The currents in each branch and their directions are labelled as shown in fig (b) above. Note that the choice of current directions depends on the battery polarity, which by convention is taken as flowing from the positive battery terminal. The values of each of the currents are determined as follows:

 R in parallel with r₂ gives an equivalent resistance of

From the equivalent circuit as shown in fig (c)

From fig (b), and using the current division rule

• Redraw the original circuit with E₁ removed, and replaced by r₁ only as shown below:

• The currents in each branch and their directions are labelled and their values determined. Resistance r₁ in parallel with R gives an equivalent resistance of:

From the equivalent circuit fig (e)

From fig (d), using the current division rule

• Fig (d) is superimposed onto fig (e) as shown below:

• The algebraic sum of the currents flowing in each branch is determined.

Resultant current flowing through source 1 is:

I₁ – I₆ = I.429 – 0. 571 = 0.858 A discharging

Resultant current flowing through source 2 is:

I₄ – I₃ = 0.857 – 1.143 = -0.286 A charging

Resultant current flowing through resistor R is:

I₂ + I₅ = 0.286 + 0.286 = 0.572 A

Thevenin’s Theorem

Sometimes to simplify the analysis of more complex circuits we can replace voltage sources and resistor networks with an equivalent voltage source and series resistor. This is called a Thevenin equivalent of the circuit.

Thevenin’s theorem states that, given a pair of terminals in a linear network, the network may be replaced by an ideal voltage source V_OC in series with resistance R_TH. V_OC is equal to the open circuit voltage across the terminals, and R_TH is the equivalent resistance across the terminals when independent voltage sources are shorted and independent current sources are replaced with open circuits.

Let’s consider the circuit below:

The part of the circuit in the fig (h) above will be replaced by its Thevenin equivalent. The open circuit voltage V_OC is found by disconnecting the rest of the circuit and determining the voltage across the terminals of the remaining open circuit.

Applying the voltage divider rule:

To find R_TH, the supply V_s is shorted i.e. V_s = 0 grounding the left end of R₁. If there were current sources in the circuit, they would be replaced with open circuits. Because R₁ and R₂ are in parallel relative to the open to the open terminals, the equivalent resistance is:

The Thevenin equivalent circuit is shown below:

The procedure adopted when using Thevenin’s theorem can be summarized as below. To determine the current in any branch of an active network i.e. one containing a source of emf:

• Remove the resistance R from that branch
• Determine the open-circuit voltage E₁ across the break
• Remove each source of emf and replace them by their internal resistances and then determine the resistance r, ‘’looking in” at the break.
• Determine the value of the current of the equivalent circuit for example for the circuit below

Worked Example

For the circuit shown below, determine the current in the 0.8 ꭥ resistor using Thevenin’s theorem.

Solution

• The 0.8 ꭥ resistor is removed from the circuit

• Calculating Current I₁

Voltage across the 4 ꭥ resistor = 4I₁ = 4 x 1.2 = 4.8 V

Voltage across AB, that is the open circuit voltage across the AB, E = 4.8 V

• Removing the source of emf gives the circuit below:

From this, we can calculate resistance r as follows:

• The equivalent Thevenin circuit is illustrated below:

The current I = 1.5 A current in the 0.8 ꭥ resistor

Norton’s Theorem

Here the linear network is replaced by an ideal current source I_SC and the Thevenin resistance R_TH in parallel with this source. I_SC is found by calculating the current that would flow through terminals if they were shorted together have removed the remaining load circuit.

It can be shown that the current I_SC flowing through R_TH produces the Thevenin voltage V_OC.

In other words, the Norton’s theorem states: The current that flows in any branch of a network is the same as that which would flow in the branch if it were connected across, a source of electrical energy, the short-circuit current of which is equal to the current that would flow in a short-circuit across the branch, and the internal resistance of which is equal to the resistance which appears across the open-circuited branch terminals.

Summary of the Procedure adopted when using Norton’s Theorem

To determine the current flowing in a resistance R of a branch AB of an active network:

• Short-circuit AB
• Determine the short-circuit I_SC flowing in the branch
• Remove all sources of emf and replace them by internal resistances (or if a current source exists, replace with an open circuit), then determine the resistance r, ‘’looking-in’’ at a break between A and B.
• Determine the current I flowing in resistance R from the Norton equivalent network (In this case with the help of current division rule).

Worked Example

Using Norton’s Theorem to determine the current flowing in the 10 ꭥ resistance for the circuit below:

Solution

• The branch containing the 10 ꭥ resistance is short-circuited as shown below:

• Circuit below is the equivalent of Fig N1

• If the 10 V source of emf is removed from the fig N1, the resistance ‘’looking’’ in at a break made between A and B is given by:

• The Norton equivalent network is shown in the fig N3

The current in the 10 ꭥ resistance can be found using current division rule by:

Maximum Power Transfer Theorem

The maximum power transfer theorem states that: The power transferred from a supply source to a load is at its maximum when the resistance of the load is equal to the internal resistance of the source.

This applies to DC as well as AC Power.

Worked Example

An audio amplifier produces an alternating output of 12 V before the connection to a load. The amplifier has an equivalent resistance of 15 ꭥ at the output. What resistance the load need to have to produce maximum power? Also, calculate the power output under this condition.

Solution

In order to produce maximum power, the load e.g. a speaker should have a resistance of 15 ꭥ to match the amplifier as shown in the circuit below:

Hence the load required, R_L = 15ꭥ

Power delivered to load, P = I²R_L = 0.4² x 15 = 2.4 W

Applications of Maximum Power Transfer Theorem

Typical applications of the maximum power transfer theorem are found in stereo amplifier design, seeking to maximize power delivered to speakers and in electric vehicle design, seeking to maximize the power delivered to driver motor.