Electronics/DC Circuit Analysis

DC Circuit Analysis
In this chapter, capacitors and inductors will be introduced (without considering the effects of AC current.) The big thing to understand about Capacitors and Inductors in DC Circuits is that they have a transient (temporary) response. During the transient period, capacitors build up charge and stop the flow of current (eventually acting like infinite resistors.) Inductors build up energy in the form of magnetic fields, and become more conductive. In other words, in the steady-state (long term behavior), capacitors become open circuits and inductors become short circuits. Thus, for DC analysis, you can replace a capacitor with an empty space and an inductor with a wire. The only circuit components that remain are voltage sources, current sources, and resistors.

Capacitors and Inductors at DC


DC steady-state (meaning the circuit has been in the same state for a long time), we've seen that capacitors act like open circuits and inductors act like shorts. The above figures show the process of replacing these circuit devices with their DC equivalents. In this case, all that remains is a voltage source and a lone resistor. (An AC analysis of this circuit can be found in the AC section.)

Resistors
If a circuits contains only resistors possibly in a combination of parallel and series connections then an equivalent resistance is determined. Then Ohm's Law is used to determine the current flowing in the main circuit. A combination of voltage and current divider rules are then used to solve for other required currents and voltages.

Simplify the following: (a) (b) (c)
 * Figure 1: Simple circuits series circuits.

The circuit in Figure 1 (a) is very simple if we are given R and V, the voltage of the source, then we use Ohm's Law to solve for the current. In Figure 1 (b) if we are given R1, R2 and V then we combine the resistor into an equivalent resistors noting that are in series. Then we solve for the current as before using Ohm's Law. In Figure 1 (c) if the resistors are labeled clockwise from the top resistor R1, R2 and R3 and the voltage source has the value: V. The analysis proceeds as follows.
 * $$R_{eq}=R_1+R_2+R_3 \,$$

This is the formula for calculating the equivalent resistance of series resistor. The current is now calculated using Ohm's Law.


 * $$i=\frac {V }{R_{eq}}= \frac {V}{R_1+R_2+R_3}$$

If the voltage is required across the third resistor then we can use voltage divider rule.
 * $$V_{R3}=\frac {VR_3}{R_1+R_2+R_3}$$

Or alternatively one could use Ohm's Law together with the current just calculated.
 * $$V_{R3}=iR_3= \frac {VR_3}{R_1+R_2+R_3}$$

(a) (b)
 * Figure 2: Simple parallel circuits.

In Figure 2 (a) if the Resistor nearest the voltage source is R1 and the other resistor R2. If we need to solve for the current i. then we proceed as before. First we calculate the equivalent resistance then use Ohm's Law to solve for the current. The resistance of a parallel combination is:
 * $$R_{eq}=\frac {R_1R_2}{R_1+R_2}$$

So the current, i, flowing in the circuit is, by Ohm's Law:
 * $$i=\frac {V(R_1+R_2)}{R_1R_2}$$

If we need to solve for current through R2 then we can use current divider rule.
 * $$i_{R2}=\frac {V(R_1+R_2)(R_1)}{(R_1R_2)(R_1+R_2)}= \frac {V}{R_2}$$(1)

But it would probably have been simpler to have used the fact that V most be dropped across R2. This means that we can simply use Ohm's Law to calculate the current through R2. The equation is just equation 1. In Figure 2 (b) we do exactly the same thing except this time there are three resistors this means that the equivalent resistance will be:
 * $$ \frac {1}{R_{eq}}= \frac {1}{R_1}+\frac {1}{R_2}+\frac {1}{R_3}$$
 * $$R_{eq} =\frac {R_1R_2R_3}{R_1R_2+R_1R_3+R_2R_3}$$

Using this fact we do exactly the same thing.

(a) (b) (c)
 * Figure 3: Combined parallel and series circuits

In Figure 3 (a), if the three resistors in the outer loop of the circuit are R1, R2 and R3 and the other resistor is R4. It is simpler to see what is going on if we combine R2 and R3 into their series equivalent resistance $$R_23$$. It is clear now that the equivalent resistance is R1 in series with the parallel combination of $$R_23$$ and R4. If we want to calculate the voltage across the parallel combination of R4 and $$R_{23}$$ then we just use voltage divider.
 * $$V_{4 | | 23}=\frac {R_4 | |(R_2+R_3)V}{R_4 | |(R_2+R_3) + R_1}$$

If we want to calculate the current through R2 and R3 then we can use the voltage across $$R_4 | |(R_2+R_3)$$ and Ohm's law.
 * $$i_{23} = \frac {R_4 | |(R_2+R_3)V}{(R_4 | |(R_2+R_3) + R_1)(R_2+R_3)}$$

Or we could calculate the current in the main circuit and then use current divider rule to get the current.

In Figure 3 (b) we take the same approach simplifying parallel combinations and series combinations of resistors until we get the equivalent resistance.

In Figure 3 (c) this process doesn't work then because there are resistors connected in a delta this means that there is no way to simplify this beyond transforming them to a star or wye connection.

Note: To calculate the current draw from the source the equivalent resistance always must be calculated. But if we just need the voltage across a series resistor this may be necessary. If we want to calculate the current in parallel combination then we must use either current divider rule or calculate the voltage across the resistor and then use Ohm's law to get the current. The second method will often require less work since the current flowing from the source is required for the use of current divider rule. The use of current divider rule is much simpler in the case when the source is a current source because the value of the current is set by the current source.



The above image shows three points 1, 2, and 3 connected with resistors R1, R2, and R3 with a common point. Such a configuration is called a star network or a Y-connection.



The above image shows three points 1, 2, and 3 connected with resistor R12, R23, and R31. The configuration is called a delta network or delta connection.

We have seen that the series and parallel networks can be reduced by the use of simple equations. Now we will derive similar relations to convert a star network to delta and vice versa. Consider the points 1 and 2. The resistance between them in the star case is simply

R1 + R2

For the delta case, we have

R12 || (R31 + R23) 

We have similar relations for the points 2, 3 and 3, 1. Making the substitution r1= R23 etc., we have, simplifying,

$$ R_1 = \frac{r_2r_3}{r_1 + r_2 + r_3} $$

$$ r_1 = \frac{R_1R_2 + R_2R_3 + R_3R_1}{R_1} $$

in the most general case. If all the resistances are equal, then R = r/3.