With connection in series, active two-terminal circuits (for example: generators), or passive (for example: resistors), are connected one after another.
Image #1: Two resistors connected in a serie.
Qualities of Connection in Series.
If we connect two resistors with connection in series to a power generator, the result of reading amperage before, in between & after resistors is always the same.
With connection in series, the amperage in each of two-terminal circuits is always the same. |
If we measure voltage on power generator's clams & on each of resitors as well, we'll notice that voltages on each of resistors are lesser than voltage of a power generator.
With a connection in series, on each of the resistors only part of voltage is lost. Whole voltage is spread across all of the resistors.
With connection in series, sum of voltage losses (voltage drops) in resistors is equal to power generator's voltage. |
Sum of all voltages in a loop shown on Image #1, with direction considered, is totalling 0 (2nd Kirchoff's Law).
This means that algebraic sum of all generators' voltages is equal to sum of all voltage losses on receivers.
If we measure (with ohmmeter) resistance of each of resistors & resistance of the whole circuit, we'll notice that resistance of whole circuit is equal to sum of resistances of each of resistors.
With connection in series, total resistance of whole circuit is equal to sum of individual resistances of each of particular resistors. |
Current flowing through total resistance of whole circuit has the same intensity (amperage) as the current flowing through resistors connected in series.
If individual resistors are equal, then with n resistors, total resistance is:
On greater resistance there's greater voltage loss.
With connection in series, individual voltage drops are proportional to appropriate resistances.
| U1 | R1 | U1 | R1 | |||
| = | = | |||||
| U | R | U2 | R2 |
U - generator's voltage;
U1, U2, ... - voltage drops;
R - total resistance;
R1, R2, ... - individual resistors;
Elements are connected in series when allowed voltage of individual element is smaller than generator's voltage.
Additional series-connected resistors.
Receiver (load) can be connected to a voltage greater than it's rated voltage if we add additional series-connected resistor(s). This resistor must be selected appropriately, to contain excess of a voltage & to survive the rated voltage of the generator.
Let's connect in series a light bulb 6V/0.3A with a generator with voltage of 24V. Let's use a resistor with appropriate resistance, to let it contain a voltage of 18V.
Image #2: A circuit with a resistor in a series-connection.
Voltage loss on leads.
In every circuit there are connected in series:
- power lead,
- receiver (load),
- return lead.
Because leads have a certain resistance Rl, with every receiver connection we have resistors connected in series.
Voltage loss on leads lowers receiver's (load's) voltage.
If we connect a light bulb to accumulator with a very long lead & if we measure voltage on a generator as well as a voltage on receiver (on load), we'll notice that a voltage on receiver (load) is lower than voltage on accumulator (generator). This is shown on image #3.
Image #3: A Practical Circuit, considering all of voltage losses.
ΔU = U1 - U2 | ΔU = I · Rl | |||
ΔU - voltage loss on leads. U1 - voltage loss on lead's beginning part. U2 - voltage loss on lead's ending part. I - current's intensity (amperage). Rl - lead's resistance. | ||||
On every lead through which the current flows, there's a voltage loss. |
If we'll connect in series to a first light bulb another a light bulb, voltage on receiver will be lower. Greater current's intensity (amperage) causes greater voltage loss in leads.
Voltage loss in leads is greater if the intensity (amperage) of the current is greater. Leads' greater resistance also causes greater voltage loss. |
Voltage loss in leads is a cause for energy loss, that is transformed into heat. That's why it should be as small as possible. It's often given as a percentage of rated voltage.
Example #1:
Let's calculate a voltage loss in leads with a resistance of Rl = 2Ω connecting a light bulb 4,5V/1A with an accumulator.
No comments:
Post a Comment