Does Voltage Change In A Series Circuit

Series Circuits

Equally mentioned in the previous section of Lesson iv, two or more than electrical devices in a excursion tin be continued by series connections or past parallel connections. When all the devices are connected using serial connections, the circuit is referred to as a serial circuit . In a series circuit, each device is continued in a mode such that there is but ane pathway past which charge can traverse the external circuit. Each charge passing through the loop of the external circuit will pass through each resistor in sequent way.

A brusque comparison and dissimilarity between serial and parallel circuits was made in the previous section of Lesson 4. In that section, it was emphasized that the human action of calculation more resistors to a series circuit results in the rather expected issue of having more than overall resistance. Since there is only one pathway through the circuit, every charge encounters the resistance of every device; so adding more than devices results in more overall resistance. This increased resistance serves to reduce the rate at which charge flows (as well known equally the current).

Equivalent Resistance and Current

Charge flows together through the external circuit at a rate that is everywhere the same. The electric current is no greater at one location as it is at some other 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 six-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 exist equivalent to having one 18-Ω resistor in the excursion. And the presence of four half-dozen-Ω resistors in series would exist equivalent to having one 24-Ω resistor in the circuit.

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

R_eq = R₁ + R₂ + R₃ + ...

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

More Do

Make, solve and check your own problems by using the Equivalent Resistance widget beneath. Make yourself a problem with any number of resistors and any values. Solve the problem; then click on the Submit button to check your respond.

The current in a series circuit is everywhere the same. Charge does Not pile up and brainstorm to accumulate at any given location such that the current at ane location is more at other locations. Accuse does NOT get used upwardly past 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 electrical circuit, everywhere marching at the same rate. Current - the rate at which accuse flows - is everywhere the same. It is the same at the first resistor as it is at the terminal resistor as information technology is in the battery. Mathematically, one might write

I_battery = I_one = I₂ = I_three = ...

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

These electric current values are hands calculated if the bombardment voltage is known and the individual resistance values are known. Using the private resistor values and the equation above, the equivalent resistance tin be calculated. And using Ohm'southward 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.

I_battery = I₁ = I_ii = I_three = ΔV_battery / R_eq

Electric Potential Difference and Voltage Drops

As discussed in Lesson 1, the electrochemical cell of a circuit supplies energy to the charge to motion it through the cell and to institute an electric potential difference across the 2 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 accuse moves through the external circuit, it encounters a loss of i.five volts of electrical potential. This loss in electrical potential is referred to as a voltage drop . It occurs every bit the electric free energy of the charge is transformed to other forms of energy (thermal, light, mechanical, etc.) within the resistors or loads. If an electric excursion powered by a 1.5-volt cell is equipped with more than i resistor, so the cumulative loss of electrical potential is 1.v volts. In that location is a voltage drop for each resistor, merely the sum of these voltage drops is 1.five volts - the same every bit the voltage rating of the power supply. This concept can be expressed mathematically by the post-obit equation:

ΔV_battery = ΔV₁ + ΔV₂ + ΔV₃ + ...

To illustrate this mathematical principle in action, 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 deviation across the light bulbs in each circuit. To determine their values, you lot would take to apply the equation above. The battery is depicted past its customary schematic symbol and its voltage is listed next to it. Make up one's mind the voltage drop for the two light bulbs and and so click the Bank check Answers button to see if y'all are correct.

Before in Lesson 1, the use of an electric potential diagram was discussed. An electrical potential diagram is a conceptual tool for representing the electric potential departure betwixt several points on an electric circuit. Consider the circuit diagram below and its respective electric potential diagram.

The circuit shown in the diagram above is powered by a 12-volt energy source. There are three resistors in the circuit connected in series, each having its ain voltage drib. The negative sign for the electric potential difference just 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 terminal to the negative terminal. Since the schematic symbol for a voltage source uses a long bar to represent the positive final, location A in the diagram is at the positive terminal or the high potential terminal. Location A is at 12 volts of electric potential and location H (the negative terminal) is at 0 volts. In passing through the battery, the accuse gains 12 volts of electric potential. And in passing through the external circuit, the charge loses 12 volts of electrical potential as depicted by the electric potential diagram shown to the right of the schematic diagram. This 12 volts of electrical potential is lost in iii steps with each stride corresponding to the flow through a resistor. In passing through the connecting wires between resistors, at that place is niggling loss in electric potential due to the fact that a wire offers relatively trivial resistance to the flow of accuse. Since locations A and B are separated by a wire, they are at well-nigh the same electric potential of 12 5. When a charge passes through its first resistor, it loses 3 Five of electric potential and drops downwardly to 9 5 at location C. Since location D is separated from location C past a mere wire, information technology is at virtually the same nine 5 electric potential as C. When a accuse passes through its second resistor, information technology loses vii V of electric potential and drops down to 2 V at location E. Since location F is separated from location E by a mere wire, it is at virtually the same ii 5 electrical potential as E. Finally, every bit a accuse passes through its last resistor, information technology loses 2 V of electric potential and drops down to 0 V at G. At locations One thousand and H, the charge is out of free energy and needs an free energy boost in order to traverse the external circuit again. The energy boost is provided by the battery as the accuse is moved from H to A.

In Lesson iii, Ohm's law (ΔV = I • R) was introduced every bit an equation that relates the voltage drib beyond a resistor to the resistance of the resistor and the current at the resistor. The Ohm's police force equation can be used for any individual resistor in a series circuit. When combining Ohm'southward police with some of the principles already discussed on this folio, a large idea emerges.

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

Since the current is everywhere the same within a serial circuit, the I value of ΔV = I • R is the same in each of the resistors of a series circuit. And so the voltage drop (ΔV) will vary with varying resistance. Wherever the resistance is greatest, the voltage drib will be greatest near that resistor. The Ohm's law equation can exist used to not only predict that resistor in a serial circuit will take the greatest voltage drop, it tin can likewise be used to calculate the actual voltage drop values.

Δ Five1 = I • Rone

Δ V2 = I • Rtwo

Δ Five3 = I • R3

Mathematical Analysis of Series Circuits

The above principles and formulae tin can be used to analyze a serial excursion and determine the values of the electric current at and electric potential difference beyond each of the resistors in a series circuit. Their utilize 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 (R_eq), the current at the battery (I_tot), and the voltage drops and current for each of the three resistors.

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

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

Now that the equivalent resistance is known, the electric current at the battery can be adamant using the Ohm's law equation. In using the Ohm's law equation (ΔV = I • R) to make up one's mind the current in the circuit, it is of import to use the battery voltage for ΔV and the equivalent resistance for R. The calculation is shown here:

I_tot = ΔV_battery / R_eq = (sixty V) / (40 Ω) = one.5 amp

The 1.v amp value for current is the current at the bombardment location. For a series circuit with no branching locations, the current is everywhere the same. The current at the battery location is the aforementioned as the current at each resistor location. Subsequently, the 1.5 amp is the value of I_one, I_ii, and I₃.

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

There are iii values left to be adamant - the voltage drops across each of the individual resistors. Ohm'southward constabulary 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 argument). The calculations are shown beneath.

ΔV1 = I1 • Ri ΔV1 = (1.5 A) • (17 Ω) ΔV1 = 25.5 Five

ΔVtwo = Iii • R2 ΔVtwo = (1.5 A) • (12 Ω) ΔV2 = 18 V

ΔViii = I3 • Riii ΔViii = (1.5 A) • (11 Ω) ΔV3 = sixteen.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 private resistor is equal to the voltage rating of the battery. In other words, is ΔV_bombardment = ΔV_ane + ΔV₂ + ΔV₃ ?

Is ΔV_battery = ΔV₁ + ΔV₂ + ΔV_iii ?

Is 60 V = 25.five Five + 18 V + 16.five V ?

Is 60 V = sixty V?

Yeah!!

The mathematical assay of this series circuit involved a blend of concepts and equations. Equally is oftentimes the instance in physics, the divorcing of concepts from equations when embarking on the solution to a physics trouble is a dangerous act. Hither, i 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 beyond each resistor in order to complete the mathematical assay. In the next part of Lesson 4, parallel circuits will exist analyzed using Ohm'south police force and parallel circuit concepts. We will meet that the approach of blending the concepts with the equations will be equally of import to that analysis.

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Bank check Your Agreement

1. Employ your understanding of equivalent resistance to complete the following statements:

a. Two three-Ω resistors placed in serial would provide a resistance that is equivalent to ane _____-Ω resistor.

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

c. Three 5-Ω resistors placed in series would provide a resistance that is equivalent to 1 _____-Ω resistor.

d. Three resistors with resistance values of 2-Ω , four-Ω , and 6-Ω are placed in series. These would provide a resistance that is equivalent to one _____-Ω resistor.

e. 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, iii-Ω, and 21-Ω are placed in series. These would provide a resistance that is equivalent to one _____-Ω resistor.

 

2. As the number of resistors in a serial excursion increases, the overall resistance __________ (increases, decreases, remains the aforementioned) and the current in the excursion __________ (increases, decreases, remains the same).

iii. Consider the post-obit two diagrams of series circuits. For each diagram, utilize arrows to indicate the direction of the conventional current. Then, make comparisons of the voltage and the current at the designated points for each diagram.

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

a. All three bulbs will have the same brightness.

b. The bulb betwixt 10 and Y will be the brightest.

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

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

5. Three identical lite bulbs are connected to a bombardment every bit shown at the right. Which adjustments could exist made to the circuit that would increment the electric current being measured at X? List all that apply.

a. Increase the resistance of one of the bulbs.

b. Increase the resistance of two of the bulbs.

c. Decrease the resistance of two of the bulbs.

d. Increase the voltage of the battery.

due east. Decrease the voltage of the battery.

f. Remove one of the bulbs.

6. Three identical lite bulbs are connected to a bombardment every bit shown at the right. W, X,Y and Z represent locations along the circuit. Which one of the following statements is truthful?

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

b. The potential deviation between Ten and Y is greater than that between Y and W.

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

d. The potential deviation between 10 and Z is greater than that betwixt Z and Westward.

east. The potential difference between X and Due west is greater than that beyond the battery.

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

7. Compare circuit X and Y below. Each is powered by a 12-volt battery. The voltage driblet across the 12-ohm resistor in circuit Y is ____ the voltage drop across the single resistor in X.

a. smaller than

b. larger than

c. the same as

8. A 12-V bombardment, a 12-ohm resistor and a light seedling are connected every bit shown in circuit Ten beneath. A half dozen-ohm resistor is added to the 12-ohm resistor and bulb to create circuit Y as shown. The seedling will announced ____.

a. dimmer in circuit Ten

b. dimmer in excursion Y

c. the aforementioned brightness in both circuits

9. Three resistors are continued in series. If placed in a excursion with a 12-volt power supply. Determine the equivalent resistance, the full excursion electric current, and the voltage driblet beyond and current at each resistor.

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

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