–  unrealisable, but useful in circuit analysis

–  can be a fixed current source, or a controlled or dependent current source

–  while an ideal voltage source has zero output resistance, an ideal current source has infinite output resistance

–  constant of proportionality is the resistance $R$
–  hence

–  current through a resistor causes power dissipation

–  each electron has a charge of:
$ 1.6 \times 10^{-19}$ coulombs

–  conventional current flows in the opposite direction

–  unit of e.m.f. is the volt

–  a volt is the potential difference

–  in analysing circuits we use ideal voltage sources

–  we also use controlled or dependent voltage sources

– a point in a circuit where two or more circuit components are joined

– any closed path that passes through no node more than once

– a loop that contains no other loop

– $\text{A, B, C, D, E and F}$ are nodes

– the paths $\text{ABEFA, BCDEB and ABCDEFA}$ are loops

– $\text{ABEFA and BCDEB}$ are meshes

–  resistance of a given sample of material is determined by its electrical characteristics and its construction

–  electrical characteristics described by its resistivity ρ or its conductivity σ (where σ = 1/ρ)

– if currents flowing into the node are positive, currents flowing out of the node are negative, then $\Sigma I=0$

– if clockwise voltage arrows are positive and anticlockwise arrows are negative then $\Sigma V=0$

–  if the output is shorted to ground, $R _2$ is in parallel with $R _3$ and the current taken from the source is $30V/15 \; k\Omega = 2mA$. This will divide equally between $R _2$ and $R _3$ so the output current, and so:
$$\large I_{\large sc} = 1mA$$

–  the resistance in the equivalent circuit is therefore given by:
$$ \large R = \frac{V_{oc}}{I_{sc}} = \frac{15V}{1mA} = 15k \Omega$$

– hence equivalent circuits are:

–  if nothing is connected across the output no current will flow in $R _2$ so there will be no voltage drop across it. Hence $V_o$ is determined by the voltage source and the potential divider formed by $R _1$ and $R _3$. Hence:
$$\large V_{\large oc} = \frac{30}{2} = 15V$$

– next consider the effect of the $20V$ source alone

– so, the output of the complete circuit is the sum of these two voltages

– next we sum the currents flowing into the nodes for which the node voltages are unknown. This gives

– solving these two equations gives

– and the required current is given by

1. Chose one node as the reference node

2. Label remaining nodes $V_1$, $V_2$ etc.

3. Label any known voltages

4. Apply Kirchhoff’s current law to each unknown node

5. Solve simultaneous equations to determine voltages

6. If necessary calculate required currents

– first we pick a reference node and label the various node voltages, assigning values where these are known

– next apply Kirchhoff’s law to each loop. This gives
\begin{align*} E - V_A-V_C-V_F-V_H=& \;0\\ V_C-V_B-V_D+V_E=& \;0 \\ V_F-V_E-V_G-V_J=& \;0 \end{align*}

– which gives the following set of simultaneous equations
\begin{align*} 50 - 70I_1 - 20(I_1-I_2) -30(I_1-I_3) - 40I_1 =& \;0\\ 20(I_1- I_2) - 100I_2 -80I_2 +10(I_3-I_2)=& \;0 \\ 30(I_1-I_3)-10(I_3-I_2)-60I_3 -90I_3=& \;0 \end{align*}

– these can be rearranged to give
\begin{align*} 50 - 160I_1 + 20I_2+30 I_3 =& \;0\\ 20I_1- 210I_2 + 10I_3 =& \;0 \\ 30I_1+10I_2-190I_3=& \;0 \end{align*}

– which can be solved to give
\begin{align*} I_1 =& 326mA \\ I_2 =& 34mA \\ I_3 =& 53mA \end{align*}

– the voltage across the $10 \Omega$ resistor is therefore given by
\begin{align*} V_E =& R_E(I_3-I_2) \\ =& 10(0.053 -0.034) \\ =& 0.19 \; V \end{align*}

– since the calculated voltage is positive, the polarity is as shown by the arrow with the left hand end of the resistor more positive than the right hand end

1. Identify the meshes and assign a clockwise-flowing current to each. Label these $I_1$ $I_2$ etc.

2. Apply Kirchhoff’s voltage law to each mesh

3. Solve the simultaneous equations to determine the currents $I_1$ $I_2$ etc.

4. Use these values to obtain voltages if required

– first assign loops currents and label voltages

– nodal and mesh analysis will work in a wide range of situations but are not necessarily the simplest methods

– no simple rules

– often involves looking at the circuit and seeing
which technique seems appropriate

– can be solved ‘by hand’ or by using matrix methods

\begin{align*} \text {– e.g.} \; 50 - 160I_1 + 20I_2 + 30I_3 =& 0 \qquad\qquad \qquad\qquad\qquad\qquad\qquad\\ 20I_1 - 210I_2 + 10I_3 =& 0 \\ 30I_1 + 10I_2 - 190I_3 =& 0 \end{align*}

– can be rearranged as as
\begin{align*} \; 160I_1 - 20I_2 - 30I_3 =& 50 \qquad\qquad \qquad\qquad\qquad\qquad\qquad\\ 20I_1 - 210I_2 + 10I_3 =& 0 \\ 30I_1 + 10I_2 - 190I_3 =& 0 \end{align*}

– these equations can be expressed as \begin{equation} \left[ \begin{array}{cccc} 160 & -20 & -30 \\ 20 & -210 & 10 \\ 30 & 10 & -190 \end{array} \right] \; \left[ \begin{array}{cccc} I_1 \\ I_2 \\ I_3 \end{array} \right] \; = \left[ \begin{array}{cccc} 50 \\ 0 \\ 0 \end{array} \right] \qquad\qquad \end{equation}

– which can be solved by hand (e.g. Cramer’s rule)

– or can use automated tools