Series Circuits

As mentioned in the previous section of Lesson 4, two or more electrical devices in a circuit can be connected by series connections or by parallel connections. When all the devices are connected using series connections, the circuit is referred toas a series circuit. In a series circuit, each device is connected in a manner such that there is only one pathway by which charge can traverse the external circuit. Each charge passing through the loop of the external circuit will pass through each resistor in consecutive fashion.

A short comparison and contrast between series and parallel circuits was made in the previous section of Lesson 4. In that section, it was emphasized that the act of adding more resistors to a series circuit results in the rather expected result of having more overall resistance. Since there is only one pathway through the circuit, every charge encounters the resistance of every device; so adding more devices results in more overall resistance. This increased resistance serves to reduce the rate at which charge flows (also known as thecurrent).

 

Equivalent Resistance and Current

Charge flows together through the external circuit at a rate that is everywhere the same. The current is no greater at one location as it is at another location. The actual amount of current varies inversely with the amount of overall resistance. There is a clear relationship between the resistance of the individual resistors and the overall resistance of the collection of resistors. As far as the battery that is pumping the charge is concerned, the presence of two 6-Ω ;resistors in series would be equivalent to having one 12-Ω resistor in the circuit. The presence of three 6-Ω resistors in series would be equivalent to having one 18-Ω resistor in the circuit. And the presence of four 6-Ω resistors in series would be equivalent to having one 24-Ω resistor in the circuit.

This is the concept of equivalent resistance. The equivalent resistanceof a circuit is the amount of resistance that a single resistor would need in order to equal the overall effect of the collection of resistors that are present in the circuit. For series circuits, the mathematical formula for computing the equivalent resistance (Req) is

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Req = R1 + R2 + R3 + ...

where R1, R2, and R3 are the resistance values of the individual resistors that are connected in series.

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The current in a series circuit is everywhere the same. Charge does NOT pile up and begin to accumulate at any given location such that the current at one location is more than at other locations. Charge does NOT become used up by resistors such that there is less of it at one location compared to another. The charges can be thought of as marching together through the wires of an electric circuit, everywhere marching at the same rate. Current - the rate at which charge flows - is everywhere the same. It is the same at the first resistor as it is at the last resistor as it is in the battery. Mathematically, one might write

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Ibattery = I1 = I2 = I3 = ...

where I1, I2, and I3 are the current values at the individual resistor locations.

These current values are easily calculated if the battery voltage is known and the individual resistance values are known. Using the individual resistor values and the equation above, the equivalent resistance can be calculated. And using Ohm's law (ΔV = I • R), the current in the battery and thus through every resistor can be determined by finding the ratio of the battery voltage and the equivalent resistance.

Ibattery = I1 = I2 = I3 = ΔVbattery / Req

 

 

Electric Potential Difference and Voltage Drops

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As discussed in Lesson 1, the electrochemical cell of a circuit supplies energy to the charge to move it through the cell and to establish an electric potential difference across the two ends of the external circuit. A 1.5-volt cell will establish an electric potential difference across the external circuit of 1.5 volts. This is to say that the electric potential at the positive terminal is 1.5 volts greater than at the negative terminal. As charge moves through the external circuit, it encounters a loss of 1.5 volts of electric potential. This loss in electric potential is referred to as a voltage drop. It occurs as the electrical energy of the charge is transformed to other forms of energy (thermal, light, mechanical, etc.) within the resistors or loads. If an electric circuit powered by a 1.5-volt cell is equipped with more than one resistor, then the cumulative loss of electric potential is 1.5 volts. There is a voltage drop for each resistor, but the sum of these voltage drops is 1.5 volts - the same as the voltage rating of the power supply. This concept can be expressed mathematically by the following equation:

ΔVbattery = ΔV1 + ΔV2 + ΔV3 + ...

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Mathematical Analysis of Series Circuits

The above principles and formulae can be used to analyze a series circuit and determine the values of the current at and electric potential difference across each of the resistors in a series circuit. Their use will be demonstrated by the mathematical analysis of the circuit shown below. The goal is to use the formulae to determine the equivalent resistance of the circuit (Req), the current at the battery (Itot), and the voltage drops and current for each of the three resistors.

 

The analysis begins by using the resistance values for the individual resistors in order to determine the equivalent resistance of the circuit.

Req = R1 + R2 + R3 = 17 Ω + 12 Ω + 11 Ω = 40 Ω

Now that the equivalent resistance is known, the current at the battery can be determined using the Ohm's law equation. In using the Ohm's law equation (ΔV = I • R) to determine the current in the circuit, it is important to use the battery voltage for ΔV and the equivalent resistance for R. The calculation is shown here:

Itot = ΔVbattery / Req = (60 V) / (40 Ω) = 1.5 amp

The 1.5 amp value for current is the current at the battery location. For a series circuit with no branching locations, the current is everywhere the same. The current at the battery location is the same as the current at each resistor location. Subsequently, the 1.5 amp is the value of I1, I2, and I3.

Ibattery = I1 = I2 = I3 = 1.5 amp

There are three values left to be determined - the voltage drops across each of the individual resistors. Ohm's law is used once more to determine the voltage drops for each resistor - it is simply the product of the current at each resistor (calculated above as 1.5 amp) and the resistance of each resistor (given in the problem statement). The calculations are shown below.

ΔV1 = I1 • R1

ΔV1 = (1.5 A) • (17 Ω)

ΔV1 = 25.5 V

ΔV2 = I2 • R2

ΔV2 = (1.5 A) • (12 Ω)

ΔV2 = 18 V

ΔV3 = I3 • R3

ΔV3 = (1.5 A) • (11 Ω)

ΔV3 = 16.5 V

As a check of the accuracy of the mathematics performed, it is wise to see if the calculated values satisfy the principle that the sum of the voltage drops for each individual resistor is equal to the voltage rating of the battery. In other words, is ΔVbattery = ΔV1 + ΔV2 + ΔV3 ?

Is ΔVbattery = ΔV1 + ΔV2 + ΔV3 ?

Is 60 V = 25.5 V + 18 V + 16.5 V ?

Is 60 V = 60 V?

Yes!!

 

The mathematical analysis of this series circuit involved a blend of concepts and equations. As is often the case in physics, the divorcing of concepts from equations when embarking on the solution to a physics problem is a dangerous act. Here, one must consider the concepts that the current is everywhere the same and that the battery voltage is equivalent to the sum of the voltage drops across each resistor in order to complete the mathematical analysis. In the next part of Lesson 4, parallel circuits will be analyzed using Ohm's law and parallel circuit concepts. We will see that the approach of blending the concepts with the equations will be equally important to that analysis.

 

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