Basic Electronic Circuit Relations

Series Circuits

A series circuit is a circuit where there is only one path from the source through all of the loads and back to the source. This means that all of the current in the circuit must flow through all of the loads.
One example of a series circuit is a string of old Christmas lights. There is only one path for the current to flow. Opening or breaking a series circuit such as this at any point in its path causes the entire circuit to "open" or stop operating.


Figure 1

That is because the basic requirement for the circuit to operate-a continuous, closed loop path-is no longer met. This is the main disadvantage of a series circuit. If any one of the light bulbs or loads burns out or is removed, the entire circuit stops operating. Many of today's circuits are actually a combination of elements in series and parallel to minimize the inconvenience of a pure series circuit.
Formula for Series and Parallel Connected Resistor
Let's take a closer look at how a series circuit operates and the way resistance affects the current flow.

Figure 2
 Consider a simple series circuit consisting of a 120 volt outlet as the source, a switch and a 60 watt light bulb. When the switch is open, the light cannot operate since the circuit is not complete. There is no closed-loop path for the current to flow through the circuit. When the switch is closed, the light bulb operates since the current flows through the circuit. The bulb glows at its full brightness since it receives its full 120 volts and has the design current flow (Figure 2).

If two light bulbs are connected to the circuit in series, the resistance of the circuit doubles (Figure 3). The current flow is now half of what it was when only one lamp was in the circuit as before. The voltage across each lamp is now 60 volts due to the reduced current flow. Each bulb is operating at only one-half its intended voltage, which will reduce its brightness. Since each bulb is the same size, they each see equal voltage drop.

 


Figure 3

Figure 4

If we add a third 60-watt bulb to the circuit, then each bulb will receive a third of the total circuit voltage, or 40 volts (Figure 4). Each bulb will produce even less light than before because we continue to add more resistance to the circuit each time we add a bulb.
If R1 is the resistance of the 40 watts bulb, the current I in Figure 2, 3 and 4 can be calculated by Ohm's Law and series connected resistors.
I = V / ( R #1 ) in Figure 2 I = V / ( R #2 ) in Figure 3 I = V / ( R #3 ) in Figure 4 Where V = 120 valts is the outlet voltage.
The brightness of the bulb is proportional to the power W in watts.
W = I x V ( 1 watt = 1 amp x 1 volt )
Using Ohm's Law,
V = I x R
We can write
W = I2 x R
Thus, in Figure 3, the voltage across each bulb is reduced in half ( because I = 120 / 2R, is reduced in half ), and the power output ( i.e., the brightness ) is reduced by a factor of 4 ( W = I2 x R. Since R does not change and I is reduced in half, W = (1/2) 2 R = R / 4 ).


Series and Parallel Circuits
Series Circuits
Parallel Circuits
Combinations



Go to TOPIC 4. Electronic Laboratory

Back to Description