SERIES-PARALLEL CIRCUITS
SERIES CONNECTION OF RESISTORS
PARALLEL CONNECTION OF RESISTORS
SERIES CONNECTION OF CAPACITORS
PARALLEL CONNECTION OF CAPACITORS
SERIES CONNECTION OF INDUCTORS
PARALLEL CONNECTION OF INDUCTORS
This chapter describes the elementary connection patterns for resistors, capacitors and inductors in a circuit.
SERIES CONNECTION OF RESISTORS
A circuit with three resistors in series is shown in Fig. 1. We need to use Kirchoff's voltage law. It is necessary to mark the loop current, and assign polarity to resistor voltages, as shown in Fig. 2.
Using Ohms' law and Kirchoff's voltage law, we get that
The above equation can be expressed as follows:
It is seen that the three resistors connected in series can be replaced by a single equivalent resistor and its resistance equals the sum of the three resistances. This rule can be extended to any number of resistors in series.
From Fig. 2, we find also that
From equation (4.3), we can generalize that when resistors are connected in series, voltage division among them is proportionate to the resistance, the scale factor relating the voltage to the resistance being the current through it. That is,
A circuit
with three resistors in parallel is shown in Fig. 3. Mark the source voltage, and the currents through the resistors, as shown in Fig. 4.
Using KCL, we obtain that
Let the equivalent resistance of three resistors in parallel be Req. If Req is connected across source Is shown in Fig. 4, the voltage across the source will be Vs. Then equation (4.5) can be presented as follows:
On cancelling the common numerator term, we obtain that
The reciprocal of resistance is conductance. Usually the letter G is used to denote conductance and its unit is siemens(S). Let
Then
That is,
When resistors are in parallel, the equivalent resistance can be found out as shown in equation (4.10). This method can be extended to any number of resistors in parallel.
The voltage across each of the resistors in Fig. 4.2 is the same. That is,
Then
It is seen from equation (4.12) that when resistors are in parallel, current division is proportionate to conductance and the scale factor relating current to conductance is the source voltage.
Many circuits have combined connections. The circuit in Fig. 5 has a combined connection. It is of interest to find out the resistance seen by the source. To calculate the resistance seen by the source, the circuit in Fig. 5 can be re-drawn as shown in Fig. 6, where Rp equals R2 and R3 connected in parallel.
Then
Capacitors can also be connected in series or parallel, in order to meet the performance requirements. In Fig. 4, three capacitors are in connected in series. Since they are in series, the current through them is the same, and each capacitor holds the same charge. From Fig. 7,
Since the charge held by a capacitor equals the product of the voltage across it and its capacitance,
It can be stated that an equivalent capacitance, Ceq has the charge Q and a voltage of E across it. Then
By connecting capacitors in series, as shown in Fig. 8,, the resultant equivalent capacitance obtained is smaller in value, but it can sustain a much higher voltage across it. Hence capacitors are connected in series in order to get an equivalent capacitor of a much higher voltage rating. In practice, in order to ensure that the capacitors share the voltage in the desired proportion, a relatively large valued resistor is connected across each capacitor. The resistor connected across each capacitor has to be large in order to reduce the power loss that can occur.
PARALLEL CONNECTION OF CAPACITORS
The circuit in Fig. 9 shows three capacitors in parallel. The total charge held by the three capacitors is the sum of the charges held by those capacitors. Let the total charge be QT. Then
It can be said that an equivalent capacitance Ceq has charge QT, with its voltage equal to E. Then
QT = Ceq × E.
That is,
Capacitors are connected in parallel in order to get an equivalent capacitance of a much higher value. It is possible that a single capacitor with the desired capacitance may not be available and hence capacitors may be connected in parallel. In some power supply filter circuits, it may be desirable to connect capacitors in parallel in order to ensure that these capacitors have the adequate ripple current rating. For example, a 1000 mF, 100 V may have a ripple current rating of 5 A, whereas two 500 mF, 100 V may have a combined ripple current rating of 6 A and hence it may be preferable to connect them in parallel.
It is possible to connect inductors in series, as shown in Fig. 11. Assume that the source voltage varies as a function of time. Then the current through the inductors would change. Since they are connected in series, the rate of change of current is the same for all the inductors. Then
It can be seen that the three inductors in series is equivalent to one inductor, with its inductance being equal to the sum of inductances. In practice, connecting inductors in series may be a convenient method to obtain a larger inductor.
Three inductors are connected in parallel, as shown in Fig. 12. It can be seen that
Let there be an equivalent inductance such that its rate of change of current equals diS/dt when the voltage across it equals vS(t). Then
This page has illustrated how resistors, inductos and capacitors in series or parallel can be replaced by an equivalent component. It is also shown how the resistors in series divide the voltage applied to the string of resistors in series. Current division among resistors in parallel has also been explained. Next chapter describes the theorems used in circuit analysis.
E4.1:
For the circuit in Fig. EP1, obtain the current drawn from the source. Find also currents through each of the resistors.
Find the power delivered by the source, and the power consumed by each of the resistors. Verify that the algebraic sum of power in all elements is zero.
E4.2
For the circuit in Fig. EP2, obtain the voltage across the source. Find also currents through each of the resistors.
E4.3:
For the circuit in Fig .EP3, find the value of resistance, Rab.
E4.4: For the circuit in Fig .EP4, find the value of resistance, Rab.
E4.5: For the circuit in Fig. EP5, find the loop current, I, and the voltage, Va.
