what resistors are considered in series to each other

Serial Circuits

As mentioned in the previous section of Lesson 4, two or more electrical devices in a excursion tin can be connected by series connections or by parallel connections. When all the devices are connected using series connections, the excursion is referred to equally a serial circuit . In a series excursion, each device is connected in a style 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 laissez passer through each resistor in consecutive fashion.

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

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 equally it is at some other location. The actual corporeality of electric current varies inversely with the corporeality of overall resistance. There is a articulate relationship between the resistance of the individual resistors and the overall resistance of the collection of resistors. Equally far as the battery that is pumping the charge is concerned, the presence of two half-dozen-Ω ;resistors in serial 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 excursion. And the presence of four half dozen-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 resistance of a excursion 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 (R_eq) is

R_eq = R₁ + R₂ + R₃ + ...

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

More than Practise

Brand, solve and cheque your own problems past using the Equivalent Resistance widget beneath. Make yourself a problem with whatsoever number of resistors and any values. Solve the problem; then click on the Submit button to cheque your answer.

The current in a series excursion is everywhere the same. Accuse does NOT pile upwards and brainstorm to accumulate at whatever given location such that the current at one location is more than than at other locations. Charge does Not go used up by resistors such that there is less of it at one location compared to some other. The charges can be thought of as marching together through the wires of an electrical circuit, everywhere marching at the same rate. Electric current - the rate at which accuse flows - is everywhere the aforementioned. It is the aforementioned at the first resistor as it is at the concluding resistor as it is in the bombardment. Mathematically, one might write

I_battery = 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 higher up, 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 past finding the ratio of the battery voltage and the equivalent resistance.

I_battery = I₁ = I_ii = I₃ = ΔV_battery / R_eq

Electrical Potential Departure and Voltage Drops

As discussed in Lesson 1, the electrochemical cell of a circuit supplies energy to the charge to motility information technology through the cell and to establish an electrical potential difference beyond the two ends of the external circuit. A 1.five-volt prison cell will plant an electrical potential difference across the external circuit of 1.5 volts. This is to say that the electric potential at the positive last is 1.v volts greater than at the negative terminal. As charge moves through the external circuit, information technology encounters a loss of one.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 free energy (thermal, low-cal, 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 one.5 volts. In that location 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 ability supply. This concept can exist expressed mathematically by the following equation:

ΔV_bombardment = ΔV₁ + ΔV₂ + ΔV_three + ...

To illustrate this mathematical principle in activeness, consider the two circuits shown below in Diagrams A and B. Suppose that you were to asked to decide the two unknown values of the electric potential difference across the lite bulbs in each circuit. To make up one's mind their values, you would take to use the equation above. The battery is depicted by its customary schematic symbol and its voltage is listed next to it. Make up one's mind the voltage drop for the ii light bulbs and so click the Check Answers push to meet if y'all are correct.

Earlier in Lesson i, the use of an electrical potential diagram was discussed. An electrical potential diagram is a conceptual tool for representing the electric potential divergence between several points on an electric circuit. Consider the circuit diagram below and its corresponding electrical potential diagram.

The circuit shown in the diagram higher up is powered by a 12-volt energy source. At that place are three resistors in the excursion connected in serial, each having its ain voltage drop. The negative sign for the electric potential difference simply denotes that at that place is a loss in electric potential when passing through the resistor. Conventional current is directed through the external circuit from the positive concluding to the negative terminal. Since the schematic symbol for a voltage source uses a long bar to represent the positive concluding, location A in the diagram is at the positive terminal or the high potential terminal. Location A is at 12 volts of electrical potential and location H (the negative terminal) is at 0 volts. In passing through the battery, the charge gains 12 volts of electric potential. And in passing through the external excursion, the charge loses 12 volts of electric potential as depicted by the electric potential diagram shown to the correct of the schematic diagram. This 12 volts of electrical potential is lost in 3 steps with each stride corresponding to the flow through a resistor. In passing through the connecting wires between resistors, there is niggling loss in electric potential due to the fact that a wire offers relatively little resistance to the flow of charge. Since locations A and B are separated by a wire, they are at virtually the same electric potential of 12 Five. When a charge passes through its first resistor, information technology loses iii V of electric potential and drops downwards to 9 V at location C. Since location D is separated from location C past a mere wire, information technology is at virtually the aforementioned 9 V electric potential equally C. When a charge passes through its 2d resistor, it loses 7 Five of electrical potential and drops down to 2 V at location E. Since location F is separated from location East by a mere wire, it is at virtually the aforementioned two Five electrical potential as E. Finally, as a charge passes through its last resistor, it loses 2 Five of electric potential and drops down to 0 Five at G. At locations G and H, the accuse is out of energy and needs an energy boost in club to traverse the external circuit over again. The energy boost is provided by the bombardment equally the accuse is moved from H to A.

In Lesson three, Ohm'southward law (ΔV = I • R) was introduced as an equation that relates the voltage drop beyond a resistor to the resistance of the resistor and the current at the resistor. The Ohm's law equation can be used for any individual resistor in a series circuit. When combining Ohm'due south law with some of the principles already discussed on this page, a big thought emerges.

In series circuits, the resistor with the greatest resistance has the greatest voltage drop.

Since the current is everywhere the aforementioned within a series circuit, the I value of ΔV = I • R is the same in each of the resistors of a series excursion. So the voltage drop (ΔV) volition vary with varying resistance. Wherever the resistance is greatest, the voltage drop will be greatest about that resistor. The Ohm's law equation can be used to not only predict that resistor in a series circuit will have the greatest voltage drop, it can also exist used to calculate the actual voltage drop values.

Δ Five1 = I • R1

Δ 52 = I • R2

Δ Five3 = I • R3

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 serial excursion. Their use volition be demonstrated past the mathematical analysis of the excursion shown beneath. The goal is to use the formulae to determine the equivalent resistance of the circuit (R_eq), the current at the battery (I_tot), and the voltage drops and current for each of the three resistors.

The analysis begins past using the resistance values for the private resistors in guild to determine the equivalent resistance of the excursion.

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

Now that the equivalent resistance is known, the current at the battery tin be determined using the Ohm's law equation. In using the Ohm'due south law equation (ΔV = I • R) to determine the electric 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:

I_tot = ΔV_battery / R_eq = (threescore V) / (40 Ω) = ane.5 amp

The one.v amp value for current is the electric current at the battery location. For a series circuit with no branching locations, the current is everywhere the aforementioned. The electric current at the battery location is the same as the current at each resistor location. Subsequently, the ane.v amp is the value of I_one, I₂, and I₃.

I_bombardment = I₁ = I_ii = I₃ = 1.five amp

In that location are three values left to be determined - the voltage drops across each of the individual resistors. Ohm's law is used over again to determine the voltage drops for each resistor - it is simply the production of the electric current at each resistor (calculated above every bit one.five amp) and the resistance of each resistor (given in the problem statement). The calculations are shown below.

ΔVi = Ione • R1 ΔV1 = (1.5 A) • (17 Ω) ΔV1 = 25.v V

ΔVii = I2 • Rtwo ΔV2 = (1.five A) • (12 Ω) ΔV2 = 18 V

ΔV3 = Iiii • R3 ΔV3 = (i.5 A) • (eleven Ω) ΔV3 = 16.5 V

As a check of the accuracy of the mathematics performed, it is wise to come across 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 ΔV_battery = ΔV₁ + ΔV₂ + ΔV₃ ?

Is ΔV_bombardment = ΔV₁ + ΔV₂ + ΔV₃ ?

Is 60 Five = 25.5 V + 18 V + sixteen.5 5 ?

Is sixty V = 60 V?

Yeah!!

The mathematical analysis of this series circuit involved a alloy of concepts and equations. As is often the instance in physics, the divorcing of concepts from equations when embarking on the solution to a physics trouble is a dangerous act. Hither, one must consider the concepts that the current is everywhere the aforementioned 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 run into that the approach of blending the concepts with the equations volition be equally of import to that assay.

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Check Your Understanding

one. Utilize your agreement of equivalent resistance to complete the following statements:

a. Ii iii-Ω resistors placed in series would provide a resistance that is equivalent to one _____-Ω resistor.

b. 3 three-Ω resistors placed in series would provide a resistance that is equivalent to 1 _____-Ω resistor.

c. Three five-Ω resistors placed in series would provide a resistance that is equivalent to one _____-Ω resistor.

d. 3 resistors with resistance values of ii-Ω , four-Ω , and vi-Ω are placed in series. These would provide a resistance that is equivalent to one _____-Ω resistor.

due east. Iii resistors with resistance values of 5-Ω , 6-Ω , and 7-Ω are placed in series. These would provide a resistance that is equivalent to one _____-Ω resistor.

f. Three resistors with resistance values of 12-Ω, 3-Ω, and 21-Ω are placed in serial. These would provide a resistance that is equivalent to one _____-Ω resistor.

 

2. Every bit the number of resistors in a series circuit increases, the overall resistance __________ (increases, decreases, remains the same) and the current in the circuit __________ (increases, decreases, remains the same).

3. Consider the post-obit two diagrams of serial circuits. For each diagram, use arrows to indicate the direction of the conventional current. And then, make comparisons of the voltage and the current at the designated points for each diagram.

4. Three identical light bulbs are connected to a D-jail cell as shown at the correct. Which i of the following statements is true?

a. All three bulbs will have the same brightness.

b. The bulb between X and Y volition exist the brightest.

c. The bulb between Y and Z will be the brightest.

d. The bulb betwixt Z and the battery volition be the brightest.

v. Three identical light bulbs are connected to a bombardment as shown at the right. Which adjustments could be made to the excursion that would increase the current being measured at X? List all that utilise.

a. Increase the resistance of 1 of the bulbs.

b. Increase the resistance of two of the bulbs.

c. Decrease the resistance of 2 of the bulbs.

d. Increase the voltage of the battery.

eastward. Decrease the voltage of the battery.

f. Remove i of the bulbs.

6. Three identical light bulbs are connected to a battery as shown at the right. W, X,Y and Z represent locations along the excursion. Which one of the post-obit statements is true?

a. The potential difference betwixt X and Y is greater than that betwixt Y and Z.

b. The potential difference between X and Y is greater than that betwixt Y and W.

c. The potential difference between Y and Z is greater than that between Y and W.

d. The potential difference betwixt Ten and Z is greater than that between Z and Westward.

eastward. The potential departure between Ten and W is greater than that across the battery.

f. The potential difference between 10 and Y is greater than that betwixt Z and W.

vii. Compare circuit X and Y below. Each is powered by a 12-volt battery. The voltage drop across the 12-ohm resistor in excursion Y is ____ the voltage drib across the unmarried resistor in Ten.

a. smaller than

b. larger than

c. the same as

eight. A 12-V battery, a 12-ohm resistor and a lite bulb are continued equally shown in circuit 10 below. A vi-ohm resistor is added to the 12-ohm resistor and seedling to create circuit Y as shown. The bulb volition appear ____.

a. dimmer in circuit X

b. dimmer in circuit Y

c. the same brightness in both circuits

nine. Iii resistors are continued in serial. If placed in a circuit with a 12-volt power supply. Make up one's mind the equivalent resistance, the total circuit current, and the voltage driblet across and current at each resistor.

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Source: https://www.physicsclassroom.com/class/circuits/Lesson-4/Series-Circuits

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