Lecture 3: DC Circuit Analysis

A network of passive elements and sources is a circuit.
Analysis: To determine currents or voltages in various elements (effects) due to various sources (cause).

Figure 4.1:

In circuit 4.1 $v=10V$ all the time.
Expected that all current (voltage) in (across) the elements is constant.
Inductor:

$$v=L\frac{di}{dt}$$ (4.1)

$\frac{di}{dt}=0$ implies $i=$ constant.
Hence $v=0$

Inductors act as a short cicuit for DC inputs. This would not be the case if I put a switch across a source.

Capacitor:

$$i=C\frac{dv}{dt}$$ (4.2)

as $\frac{dv}{dt}=0$ (expected), $i=0$.
Thus capacitor acts as open circuit for DC analysis.

Figure 4.2:

The resultant circuit will be as shown in Fig.4.2.

Analysis: To find currents in all branches, voltage across all branches.
We can use Kirchoff's law (voltage and current). For as many independent equations as number of unknown variables.
Solve the simultaneous equations, and get the result.

$V_{ab}=$ Voltage drop from 'a' to 'b'. Therefore, $V_{ab} = - V_{ba}$
$I_{ab}=$ Current in branch ab in the direction from 'a' to 'b'. Here $I_{ab} = - I_{ba}$.
Hence:

$$V_{fa}=-v$$

$$V_{ab}=I_{ab}\times 1=I_{ab}$$

$$V_{be}=I_{be}2$$

$$V_{ef}=I_{ef}1$$

$$V_{bc}=I_{bc}1$$

$$V_{cd}=I_{cd}2$$

$$V_{de}=I_{de}3$$

$$V_{ab}+V_{be}+V_{ef}=v$$

$$I_{fa}=I_{ab}$$

$$I_{ef}=I_{fa}$$

$$I_{bc}=I_{cd}$$

$$I_{cd}=I_{de}$$

$$V_{ab}+V_{bc}+V_{cd}+V_{de}+V_{ef}=v$$

$$I_{bc}+I_{be}=I_{ab}$$

Note: In a circuit with $N$ nodes, the number of branches $B$ will alwasy be $(N-1)+L$, where $L$ are maximum number of independent closed paths possible in the circuit.

Hence, we can always form $B$ equations using Ohm's law, $(N-1)$ equations using KCL, $L$ equations using KVL. Hence in total $2B$ equations can be formed, which are sufficient to solve for $2B$ variables (voltage and current in each branch).

Can we simplify the situation? Loop currents method: We do away with branch currents and define loop currents. The branch currents can be written in terms of loop currents once all the loop currents passing through the branch and their directions are known. The branch voltages can always be written using Ohm's law and branch current written in terms of loop currents. So now our objective is to find loop currents. For this we choose maximum number of independent loops (Fig.4.3) and apply KVL in them.

Figure 4.3:

If $i_1$ and $i_2$ are known, voltages across all elements can be found.
Make two independent equations:
For loop abef

$$v-i_1\times 1 -(i_1-i_2)\times 2 -i_1\times 1 =0$$ (4.3)

$$v=i_1(1+2+1)-2i_2$$ (4.4)

For loop bcde:

$$-(i_2-i_1)2-1i_2 - 2i_2 -3i_3=0$$ (4.5)

$$0=-2i_2 + i_2(2+1+2+3)$$ (4.6)

Use any technique to solve these (such as using matrices). We get:

$$i_1=\frac{2}{7}\;Amperes$$ (4.7)

$$i_2=\frac{2}{7}\; Amperes$$ (4.8)

Nodal Voltage Method

Figure 4.4: Nodal Voltage method

Considering Fig.4.4.

Independent Nodes: One of the nodes in circuit need to be considered as reference node. Hence its node potential is zero. For other nodes, nodal voltage is potential differetial w.r.t. to reference node. The nodes are called independent nodes. In general for $N$ node network, $N-1$ nodes will be independent.

At node b: $i_1+i_2-i_3=0$.
Similarly, other equations are:

$$\frac{E_a-E_b}{1}-\frac{E_c-E_b}{1}-\frac{E_b}{2}=0$$ (4.9)

$$\frac{E_a-E_b}{1}=i_{fa}$$ (4.10)

$$E_a=E_f+v$$ (4.11)

$$\frac{-E_f}{1}=i_{fa}$$ (4.12)

$$\frac{E_c-E_b}{1}=\frac{E_d-E_c}{2}$$ (4.13)

$$\frac{-E_d}{3}=\frac{E_d-E_c}{2}$$ (4.14)

These are six equations, in six unknowns. Thus can be solved for a unique solution. One can make a supernode and use KCL combining the nodes nodes a and f. We also make extra equations for potential difference between two nodes.
With supernode, no. of equations is equal to no. of independent nodes whose voltage w.r.t. reference needs to be determined.

Current Sources in Loop Current Analysis

Figure 4.5: Current sources in Loop current analysis

Using KVL for loop 1 in Fig.4.5:

$$5-i_1-v_{bg}-(i_1-i_2)1-i_13=0$$ (4.15)

The other equations are:

$$i_1-i_2=2$$ (4.16)

$$v_{bg}-i_22-i_21-i_22-(i_2-i_1)=0$$ (4.17)

The first and the third equation can be combined for taking care of $V_{bg}$. This can be done by making superloop for writing KVL.

Graph
For analysing circuits efficiently.
Loop current method
One can form a spanning tree from graph such that current sources are in links (Those elements which do not form part of the tree). Each link when added to the tree gives a loop. All voltage sources should be kept in branches of tree. For example, refer to the following two figures (Fig.4.6 , Fig.4.7)

Figure 4.6: Loop Current method

Figure 4.7: Loop current method

Node voltages

Figure 4.8: Node Voltage method

In the above figure, $V_{af}=2V$. There are five unknown node voltages in the above circuit, namely, $V_{ab}$, $V_{fe}$, $V_{bc}$, $V_{cd}$ and $V_{ge}$. Correspondingly, we have five equations. Note that we can merge $af$ into one supernode

$$V_{ab}+V_{fe}/3=0$$

$$V_{ab}-V_{bc}/2 + 2=0$$

$$V_{bc}/2-V_{cd}=0$$

$$V_{cd}-V_{de}/2=0$$

$$V_{ge}+2=0$$

The second equation follows from looking at node $b$, while the third one from doing the same at node $c$.