Lecture 3 (Simple circuits and networks)

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Peter's Index Physics Home Lecture 2 Course Index Lecture 4

An introduction to Electricity and Strength of Materials with Peter Eyland

In this lecture:
• Simple circuits are reduced to a single equivalent resistor by series and parallel substitution,
• Kirchhoff's rules for finding currents and potential differences in networks are demonstrated

Reduction by Series/parallel substitution

Example C7

For the circuit shown on the right, find the current through; the potential across; and the power dissipated in each resistor.

answer C7

Step 1: reduce the parallel combination.

The circuit now is now a series circuit

Step 2: reduce the series combination.

For the 50R

Step 3: add the known current(s) and potential(s) to the circuit, and calculate any unknowns.

The branch currents are:

The branch powers are:

Ratios in simple circuits

For any number of resistors connected in series, the potentials across each resistor are simply proportional to the size of the resistor over the sum of them.

Take the case of three resistors connected in series, as shown in the diagram.

Three series resistors

Since the current is given by:

current calculation

The potential difference across resistor R₁ is given by the ratio:

potential difference as a ratio

For two resistors connected in parallel, the branch currents are proportional to the resistor in the other branch over the sum of them.

Take the case of two resistors connected in parallel, as shown in the diagram.
First find the potential difference across the parallel combination.	The branch currents are given by the ratios:

Example C8

A potential difference of 120V is applied across three resistors (20R, 30R and 10R) connected in series. Find the potential difference across the 30R.

series ratio

Example C9

A current of 1mA flows into a parallel combination of 20R and 30R resistors. Find the current in the 20R resistor.

parallel ratio

Example C10.

For the resistor combination shown in the diagram, calculate the current through the 6Ω resistor.

answer C10

Firstly, find the equivalent resistance for the three resistors in parallel.

Check: the equivalent resistance is smaller than the smallest.

Secondly, find the equivalent resistance for the new two top resistors in series.

Thirdly, calculate the branch currents using the opposite resistance over the sum.

The current in the 5Ω resistor = (20/25)×60 = 48A

The current in the 20Ω resistor = (5/25)×60 = 12A

Check: the current through the smaller resistor is larger than the current through the larger resistor.

Fourthly, the equivalent resistance of the 12|4|6Ω parallel combination has been calculated as 2Ω.

The total current through the 12|4|6Ω parallel combination has been calculated as 48A.

The potential difference across the 12|4|6Ω parallel combination is then V = 2×48 = 96V.

The current through the lowest resistor (6Ω) is then V/6 = 96/6 = 16A.

Kirchhoff's rules for network circuits

Rule 1: In electric circuit networks, charge is conserved at junctions.

This means that the total current that flows into a junction must equal the total current that flows out of that junction. Symbolically, this is written as:

Rule 2: In electric circuit networks, electrical energy is conserved around a closed circuit.

This means that the power supplied by the emf's is dissipated by the resistors. Symbolically, this is written as:

total power supplied is total power dissipated

Dividing by the current, we see that the sum of the e.m.f.'s equals the sum of the potential drops across the resistors.

Kirchhoff's procedure.

(i)	Label the salient points of the circuit.
(ii)	*Apply rule 1* by going to one junction and assuming at least one current (i₁) goes into the junction and at least one current (i₂) comes out. Make up other current directions and values as needed, e.g. i₁ - i₂. It doesn't matter if you make a mistake with a direction, it will just appear in the end as a negative value.
(iii)	Complete the current flows at all the junctions. Make sure at least one current goes into all junctions and one out, and make up current directions and values as needed. As before, wrong choices for directions just appear as negative values.
(iv)	Write the polarities on the resistors. Current flows down a potential difference so write a plus on the high side and minus on the low side.
(v)	*Apply rule 2* to a number of loops (one loop for each unknown current). When you go down in potential write a minus, and when you go up, write a plus.
(vi)	Solve the equations for the unknown currents

Example C11

For the circuit shown, find the size and direction of the current through each of the three resistors.

- - - - -

At point C, choose that current i₁ flows in from B to C.
Choose that current i₂ flows in from F to C.
Then current i₃ must flow out from C to D.

Current flows down a potential difference.
The 40Ω resistor has + on the left and - on the right.
The 10Ω resistor has + on the left and - on the right.
The 20Ω resistor has - on the left and + on the right.

There are only two independent currents.

kirchhoff's loop equations and solution for C11

Example C12

For the circuit shown, find the size and direction of the current through each of the three resistors.

kirchhoff's loop equations and solution for C12

Finding potential differences between any two points in circuit networks.

First find the currents by using Kirchhoff's rules, and then start at the first point and track the potentials to the second point.

Example C13

Find the potential difference between B and E.

- - - - -

At point B, choose that current i₁ flows in from E to B.
Choose that current i₂ flows in from C to B.
Then current i₃ = i₁+i₂, must flow out from B to A.
The currents are consistent at junction E.

The circuit has been re-drawn to show the resistor polarities (current flows down a potential difference).

The loop rule is applied for two independent loops

kirchhoff's loop equations and solution for C13

The circuit has been re-drawn to show the current values and resistor polarities

Now track from B to E

V_Bis 2V higher than V_E

Summarising:

For simple circuits, currents and potential differences can be found by series and parallel substitutions.
Kirchhoff's rules are needed for networks, i.e.

Rule 1: In electric circuit networks, charge is conserved at junctions.

Rule 2: In electric circuit networks, electrical energy is conserved around a closed circuit.

(i)	Label the salient points of the circuit.
(ii)	*Apply rule 1* by going to one junction and assuming at least one current (i₁) goes into the junction and at least one current (i₂) comes out. Make up other current directions and values as needed, e.g. i₁ - i₂. It doesn't matter if you make a mistake with a direction, it will just appear in the end as a negative value.
(iii)	Complete the current flows at all the junctions. Make sure at least one current goes into all junctions and one out, and make up current directions and values as needed. As before, wrong choices for directions just appear as negative values.
(iv)	Write the polarities on the resistors. Current flows down a potential difference so write a plus on the high side and minus on the low side.
(v)	*Apply rule 2* to a number of loops (one loop for each unknown current). When you go down in potential write a minus, and when you go up, write a plus.
(vi)	Solve the equations for the unknown currents

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Take the case of three resistors connected in series, as shown in the diagram.
Since the current is given by:	The potential difference across resistor R₁ is given by the ratio: