Can I Run Two Refrigerators on the Same Circuit

Combination Circuits

Previously in Lesson iv, it was mentioned that in that location are ii dissimilar ways to connect 2 or more electrical devices together in a circuit. They can be connected by ways of series connections or by means of parallel connections. When all the devices in a circuit are continued past serial connections, then the circuit is referred to as a series circuit. When all the devices in a circuit are connected by parallel connections, then the circuit is referred to as a parallel circuit. A 3rd blazon of circuit involves the dual use of series and parallel connections in a circuit; such circuits are referred to every bit compound circuits or combination circuits. The circuit depicted at the right is an example of the utilise of both series and parallel connections within the same excursion. In this example, light bulbs A and B are connected by parallel connections and light bulbs C and D are connected by serial connections. This is an example of a combination circuit .

When analyzing combination circuits, it is critically important to take a solid understanding of the concepts that pertain to both series circuits and parallel circuits. Since both types of connections are used in combination circuits, the concepts associated with both types of circuits employ to the respective parts of the circuit. The main concepts associated with series and parallel circuits are organized in the table below.

Series Circuits The current is the aforementioned in every resistor; this current is equal to that in the battery. The sum of the voltage drops across the individual resistors is equal to the voltage rating of the battery. The overall resistance of the collection of resistors is equal to the sum of the individual resistance values, Rtot = Rane + Rii + R3 + ...

Parallel Circuits The voltage drop is the same across each parallel branch. The sum of the current in each individual branch is equal to the current outside the branches. The equivalent or overall resistance of the drove of resistors is given past the equation i/Req = 1/Ri + 1/R2 + 1/R3 ...

Each of the above concepts has a mathematical expression. Combining the mathematical expressions of the above concepts with the Ohm's police equation (ΔV = I • R) allows one to carry a complete analysis of a combination excursion.

Assay of Combination Circuits

The bones strategy for the analysis of combination circuits involves using the meaning of equivalent resistance for parallel branches to transform the combination circuit into a series circuit. Once transformed into a serial circuit, the assay can exist conducted in the usual manner. Previously in Lesson 4, the method for determining the equivalent resistance of parallel are equal, then the total or equivalent resistance of those branches is equal to the resistance of one branch divided by the number of branches.

This method is consequent with the formula

ane / R_eq = 1 / R_ane + 1 / R₂ + 1 / R_iii + ...

where R₁, R₂, and R₃ are the resistance values of the private resistors that are connected in parallel. If the two or more than resistors found in the parallel branches do not have equal resistance, and then the above formula must be used. An instance of this method was presented in a previous section of Lesson 4.

By applying one'southward understanding of the equivalent resistance of parallel branches to a combination circuit, the combination circuit can be transformed into a series excursion. And so an agreement of the equivalent resistance of a series circuit tin can be used to determine the total resistance of the circuit. Consider the following diagrams below. Diagram A represents a combination excursion with resistors R_two and R_iii placed in parallel branches. Two 4-Ω resistors in parallel is equivalent to a resistance of 2 Ω. Thus, the two branches can be replaced by a single resistor with a resistance of 2 Ω. This is shown in Diagram B. Now that all resistors are in serial, the formula for the full resistance of series resistors can be used to determine the full resistance of this circuit: The formula for series resistance is

R_tot = R_ane + R₂ + R₃ + ...

So in Diagram B, the total resistance of the circuit is xΩ.

Once the total resistance of the circuit is determined, the assay continues using Ohm's law and voltage and resistance values to make up one's mind current values at various locations. The entire method is illustrated below with two examples.

Case 1:

The first case is the easiest case - the resistors placed in parallel have the same resistance. The goal of the assay is to make up one's mind the electric current in and the voltage drop across each resistor.

As discussed above, the outset step is to simplify the excursion past replacing the two parallel resistors with a single resistor that has an equivalent resistance. Two 8 Ω resistors in series is equivalent to a single 4 Ω resistor. Thus, the ii co-operative resistors (R₂ and R_three) can be replaced by a single resistor with a resistance of 4 Ω. This four Ω resistor is in series with R₁ and R₄. Thus, the total resistance is

R_tot = R_one + 4 Ω + R₄ = 5 Ω + iv Ω + 6 Ω

R_tot = 15 Ω

Now the Ohm'southward law equation (ΔV = I • R) can be used to determine the total current in the excursion. In doing and so, the full resistance and the full voltage (or bombardment voltage) will have to be used.

I_tot = ΔV_tot / R_tot = (60 V) / (xv Ω)

I_tot = 4 Amp

The 4 Amp electric current calculation represents the electric current at the battery location. Yet, resistors R₁ and R₄ are in series and the current in series-connected resistors is everywhere the same. Thus,

I_tot = I_one = I₄ = 4 Amp

For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. Thus, I₂ + I_three must equal four Amp. There are an infinite number of possible values of I₂ and I₃ that satisfy this equation. Since the resistance values are equal, the current values in these two resistors are too equal. Therefore, the current in resistors 2 and three are both equal to 2 Amp.

I₂ = I₃ = 2 Amp

Now that the current at each individual resistor location is known, the Ohm'due south law equation (ΔV = I • R) can be used to determine the voltage drop across each resistor. These calculations are shown beneath.

ΔV_i = I₁ • R_i = (4 Amp) • (five Ω)

ΔV₁ = 20 V

ΔV₂ = I_two • R₂ = (2 Amp) • (8 Ω)

ΔV₂ = 16 V

ΔV₃ = I₃ • R₃ = (2 Amp) • (8 Ω)

ΔV₃ = 16 V

ΔV₄ = I₄ • R_four = (4 Amp) • (six Ω)

ΔV₄ = 24 V

The analysis is at present consummate and the results are summarized in the diagram below.

Example 2:

The second instance is the more than difficult instance - the resistors placed in parallel have a unlike resistance value. The goal of the analysis is the same - to make up one's mind the electric current in and the voltage drop across each resistor.

As discussed above, the kickoff step is to simplify the circuit by replacing the two parallel resistors with a single resistor with an equivalent resistance. The equivalent resistance of a 4-Ω and 12-Ω resistor placed in parallel can be adamant using the usual formula for equivalent resistance of parallel branches:

i / R_eq = 1 / R₁ + 1 / R₂ + 1 / R₃ ...

1 / R_eq = 1 / (iv Ω) + 1 / (12 Ω)

1 / R_eq = 0.333 Ω^-1

R_eq = one / (0.333 Ω^-1)

R_eq = three.00 Ω

Based on this calculation, it tin exist said that the two branch resistors (R_two and R_iii) can be replaced by a single resistor with a resistance of 3 Ω. This 3 Ω resistor is in series with R₁ and R_iv. Thus, the total resistance is

R_tot = R₁ + 3 Ω + R₄ = five Ω + 3 Ω + eight Ω

R_tot = 16  Ω

Now the Ohm's police force equation (ΔV = I • R) can be used to determine the total electric current in the circuit. In doing so, the total resistance and the total voltage (or bombardment voltage) will have to be used.

I_tot = ΔV_tot / R_tot = (24 V) / (16 Ω)

I_tot = 1.5 Amp

The 1.v Amp current calculation represents the current at the bombardment location. Yet, resistors R₁ and R₄ are in series and the current in series-connected resistors is everywhere the same. Thus,

I_tot = I₁ = I₄ = 1.5 Amp

For parallel branches, the sum of the electric current in each private branch is equal to the current exterior the branches. Thus, I_ii + I_three must equal 1.5 Amp. There are an infinite possibilities of I₂ and I₃ values that satisfy this equation. In the previous instance, the two resistors in parallel had the identical resistance; thus the current was distributed every bit among the 2 branches. In this instance, the unequal current in the two resistors complicates the analysis. The branch with the least resistance will accept the greatest current. Determining the amount of current will need that we use the Ohm'south law equation. But to use information technology, the voltage drop across the branches must first be known. So the direction that the solution takes in this instance will be slightly different than that of the simpler case illustrated in the previous example.

To decide the voltage drib across the parallel branches, the voltage drop across the two series-connected resistors (R_one and R_four) must first be determined. The Ohm'due south police force equation (ΔV = I • R) can exist used to determine the voltage driblet beyond each resistor. These calculations are shown below.

ΔV₁ = I_i • R_one = (1.5 Amp) • (5 Ω)

ΔV₁ = seven.5 Five

ΔV₄ = I₄ • R₄ = (one.five Amp) • (8 Ω)

ΔV_four = 12 5

This circuit is powered by a 24-volt source. Thus, the cumulative voltage driblet of a accuse traversing a loop about the excursion is 24 volts. There will be a 19.five V drib (7.v V + 12 5) resulting from passage through the two serial-continued resistors (R_i and R₄). The voltage drib across the branches must be 4.5 volts to make up the difference betwixt the 24 volt full and the 19.v-volt drop across R₁ and R₄. Thus,

ΔV₂ = V₃ = 4.5 V

Knowing the voltage drop across the parallel-continued resistors (R_one and R₄) allows one to apply the Ohm'southward law equation (ΔV = I • R) to decide the electric current in the two branches.

I_ii = ΔV_two / R₂ = (4.5 V) / (4 Ω)

I₂ = 1.125 A

I_iii = ΔV₃ / R₃ = (4.5 V) / (12 Ω)

I_three = 0.375 A

The analysis is now complete and the results are summarized in the diagram below.

Developing a Strategy

The ii examples above illustrate an effective concept-centered strategy for analyzing combination circuits. The approach demanded a firm grasp of the serial and parallel concepts discussed earlier. Such analyses are often conducted in club to solve a physics problem for a specified unknown. In such situations, the unknown typically varies from problem to problem. In 1 problem, the resistor values may be given and the current in all the branches are the unknown. In another problem, the current in the battery and a few resistor values may be stated and the unknown quantity becomes the resistance of ane of the resistors. Different trouble situations will apparently crave slight alterations in the approaches. Nonetheless, every problem-solving approach will utilize the aforementioned principles utilized in budgeted the two example problems to a higher place.

The post-obit suggestions for approaching combination circuit problems are offered to the beginning student:

• If a schematic diagram is not provided, have the time to construct one. Use schematic symbols such as those shown in the instance above.
• When approaching a problem involving a combination circuit, take the time to organize yourself, writing downwards known values and equating them with a symbol such as I_tot, I_one, R₃, ΔV₂, etc. The organization scheme used in the 2 examples above is an constructive starting bespeak.
• Know and use the appropriate formulae for the equivalent resistance of series-connected and parallel-connected resistors. Employ of the wrong formulae will guarantee failure.
• Transform a combination excursion into a strictly series circuit by replacing (in your mind) the parallel section with a single resistor having a resistance value equal to the equivalent resistance of the parallel section.
• Utilize the Ohm'southward police force equation (ΔV = I • R) often and appropriately. Well-nigh answers volition exist adamant using this equation. When using it, information technology is important to substitute the appropriate values into the equation. For instance, if computing I_two, it is important to substitute the ΔV₂ and the R_ii values into the equation.

For further practice analyzing combination circuits, consider analyzing the problems in the Check Your Understanding section beneath.

We Would Like to Propose ...

Why just read about it and when y'all could be interacting with it? Interact - that'southward exactly what you do when you lot use one of The Physics Classroom'southward Interactives. We would similar to suggest that you combine the reading of this folio with the utilize of our DC Circuit Builder Interactive. You can find it in the Physics Interactives section of our website. The DC Circuit Builder provides the learner with a virtual circuit edifice kit. Y'all tin easily drag voltage sources, resistors and wires onto the workspace and adjust and connect them anyhow you wish. Voltmeters and ammeters allow you lot to measure current and voltage drops. Borer a resistor or a voltage source allows you to change the resistance or the input voltage. Information technology'southward easy. It'due south fun. And it's safe (unless you're using it in the bathtub).

Check Your Understanding

1. A combination circuit is shown in the diagram at the right. Employ the diagram to reply the post-obit questions.

a. The current at location A is _____ (greater than, equal to, less than) the current at location B.

b. The electric current at location B is _____ (greater than, equal to, less than) the current at location E.

c. The current at location One thousand is _____ (greater than, equal to, less than) the current at location F.

d. The current at location E is _____ (greater than, equal to, less than) the current at location G.

e. The current at location B is _____ (greater than, equal to, less than) the current at location F.

f. The current at location A is _____ (greater than, equal to, less than) the current at location L.

g. The current at location H is _____ (greater than, equal to, less than) the current at location I.

2. Consider the combination circuit in the diagram at the right. Utilise the diagram to answer the following questions. (Presume that the voltage drops in the wires themselves in negligibly small.)

a. The electric potential difference (voltage drop) between points B and C is _____ (greater than, equal to, less than) the electric potential deviation (voltage drop) betwixt points J and Chiliad.

b. The electric potential difference (voltage drop) between points B and K is _____ (greater than, equal to, less than) the electric potential divergence (voltage drop) between points D and I.

c. The electric potential difference (voltage drib) between points E and F is _____ (greater than, equal to, less than) the electric potential difference (voltage drop) between points Thou and H.

d. The electrical potential departure (voltage drop) betwixt points East and F is _____ (greater than, equal to, less than) the electric potential difference (voltage drop) between points D and I.

e. The electric potential deviation (voltage drop) between points J and K is _____ (greater than, equal to, less than) the electric potential difference (voltage drib) between points D and I.

f. The electric potential deviation betwixt points L and A is _____ (greater than, equal to, less than) the electric potential divergence (voltage drop) between points B and Thou.

 

three. Use the concept of equivalent resistance to decide the unknown resistance of the identified resistor that would make the circuits equivalent.

4. Analyze the following circuit and determine the values of the total resistance, total electric current, and the electric current at and voltage drops across each individual resistor.

five. Referring to the diagram in question #four, determine the ...

a. ... power rating of resistor 4.

b. ... rate at which energy is consumed by resistor iii.

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

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