how to find current in a parallel circuit

Parallel Circuits

As mentioned in a previous section of Lesson four, two or more electrical devices in a excursion can be connected by series connections or by parallel connections. When all the devices are connected using parallel connections, the circuit is referred to as a parallel excursion . In a parallel excursion, each device is placed in its own divide co-operative. The presence of branch lines ways that in that location are multiple pathways by which charge can traverse the external circuit. Each accuse passing through the loop of the external excursion will laissez passer through a single resistor nowadays in a single branch. When arriving at the branching location or node, a charge makes a choice every bit to which branch to travel through on its journeying back to the low potential concluding.

A short comparison and contrast between series and parallel circuits was fabricated in an earlier section of Lesson iv. In that section, it was emphasized that the act of adding more than resistors to a parallel circuit results in the rather unexpected outcome of having less overall resistance. Since there are multiple pathways by which accuse tin can flow, adding another resistor in a divide branch provides another pathway by which to direct charge through the main surface area of resistance inside the circuit. This decreased resistance resulting from increasing the number of branches will accept the effect of increasing the charge per unit at which accuse flows (also known every bit the current). In an effort to brand this rather unexpected result more reasonable, a tollway illustration was introduced. A tollbooth is the master location of resistance to car menstruum on a tollway. Adding additional tollbooths within their ain co-operative on a tollway will provide more pathways for cars to flow through the toll station. These boosted tollbooths will decrease the overall resistance to auto menses and increase the rate at which they menses.

Current

The charge per unit at which charge flows through a circuit is known as the current. Accuse does Non pile up and begin to accrue at any given location such that the current at one location is more at other locations. Charge does Non become used up past resistors in such a manner that there is less current at one location compared to another. In a parallel circuit, charge divides upwardly into separate branches such that at that place tin can be more electric current in 1 branch than in that location is in another. Yet, when taken as a whole, the total amount of electric current in all the branches when added together is the aforementioned every bit the corporeality of current at locations outside the branches. The rule that current is everywhere the aforementioned yet works, only with a twist. The current outside the branches is the aforementioned equally the sum of the electric current in the private branches. It is still the same corporeality of current, only separate into more than one pathway.

In equation class, this principle can exist written as

I_total = I_one + I₂ + I₃ + ...

where I_total is the total amount of current outside the branches (and in the battery) and I₁, I₂, and I₃ stand for the electric current in the individual branches of the excursion.

Throughout this unit, at that place has been an extensive reliance upon the analogy between charge flow and water flow. In one case more, nosotros will return to the analogy to illustrate how the sum of the current values in the branches is equal to the corporeality outside of the branches. The catamenia of charge in wires is coordinating to the flow of h2o in pipes. Consider the diagrams beneath in which the catamenia of water in pipes becomes divided into divide branches. At each node (branching location), the water takes two or more split up pathways. The rate at which water flows into the node (measured in gallons per minute) will be equal to the sum of the flow rates in the private branches beyond the node. Similarly, when two or more branches feed into a node, the rate at which water flows out of the node volition be equal to the sum of the flow rates in the private branches that feed into the node.

The aforementioned principle of menstruum partition applies to electric circuits. The rate at which charge flows into a node is equal to the sum of the catamenia rates in the individual branches beyond the node. This is illustrated in the examples shown below. In the examples a new circuit symbol is introduced - the alphabetic character A enclosed within a circle. This is the symbol for an ammeter - a device used to measure the current at a specific betoken. An ammeter is capable of measuring the electric current while offering negligible resistance to the flow of charge.

Diagram A displays two resistors in parallel with nodes at point A and betoken B. Charge flows into indicate A at a charge per unit of half dozen amps and divides into two pathways - 1 through resistor 1 and the other through resistor 2. The current in the branch with resistor 1 is 2 amps and the current in the co-operative with resistor 2 is 4 amps. After these two branches meet once more at betoken B to form a single line, the current once more becomes 6 amps. Thus nosotros see the principle that the electric current outside the branches is equal to the sum of the current in the private branches holds true.

I_total = I₁ + I₂

vi amps = two amps + four amps

Diagram B above may be slightly more complicated with its three resistors placed in parallel. Four nodes are identified on the diagram and labeled A, B, C and D. Charge flows into point A at a charge per unit of 12 amps and divides into two pathways - one passing through resistor 1 and the other heading towards point B (and resistors 2 and iii). The 12 amps of current is divided into a 2 amp pathway (through resistor ane) and a 10 amp pathway (heading toward point B). At point B, there is further partition of the menses into ii pathways - one through resistor 2 and the other through resistor three. The electric current of 10 amps approaching point B is divided into a half dozen-amp pathway (through resistor ii) and a four-amp pathway (through resistor 3). Thus, information technology is seen that the current values in the iii branches are 2 amps, half-dozen amps and 4 amps and that the sum of the current values in the private branches is equal to the current outside the branches.

I_total = I₁ + I₂ + I₃

12 amps = 2 amps + six amps + 4 amps

A flow analysis at points C and D can also be conducted and information technology is observed that the sum of the flow rates heading into these points is equal to the flow charge per unit that is found immediately across these points.

Equivalent Resistance

The bodily amount of current always varies inversely with the corporeality of overall resistance. There is a clear relationship betwixt the resistance of the individual resistors and the overall resistance of the drove of resistors. To explore this relationship, allow's begin with the simplest example of ii resistors placed in parallel branches, each having the same resistance value of 4 Ω. Since the circuit offers two equal pathways for accuse flow, only half the accuse will choose to laissez passer through a given branch. While each individual co-operative offers 4 Ω of resistance to whatsoever accuse that flows through it, only one-half of all the charge flowing through the excursion will meet the 4 Ω resistance of that individual co-operative. Thus, as far as the battery that is pumping the accuse is concerned, the presence of two 4-Ω resistors in parallel would be equivalent to having one 2-Ω resistor in the circuit. In the aforementioned manner, the presence of ii 6-Ω resistors in parallel would be equivalent to having one 3-Ω resistor in the circuit. And the presence of two 12-Ω resistors in parallel would be equivalent to having one vi-Ω resistor in the circuit.

Now allow's consider another simple instance of having iii resistors in parallel, each having the same resistance of 6 Ω. With 3 equal pathways for charge to menstruum through the external excursion, only one-tertiary the charge will choose to laissez passer through a given co-operative. Each individual branch offers half-dozen Ω of resistance to the accuse that passes through information technology. However, the fact that but one-third of the accuse passes through a particular branch means that the overall resistance of the circuit is 2 Ω. As far as the bombardment that is pumping the charge is concerned, the presence of three 6-Ω resistors in parallel would be equivalent to having one 2-Ω resistor in the circuit. In the same manner, the presence of iii ix-Ω resistors in parallel would be equivalent to having i three-Ω resistor in the circuit. And the presence of 3 12-Ω resistors in parallel would exist equivalent to having one 4-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 outcome of the collection of resistors that are present in the excursion. For parallel circuits, the mathematical formula for computing the equivalent resistance (R_eq) is

1 / R_eq = 1 / R₁ + 1 / R₂ + 1 / R_three + ...

where R_one, R₂, and R₃ are the resistance values of the private resistors that are continued in parallel. The examples above could be considered simple cases in which all the pathways offer the same amount of resistance to an individual charge that passes through it. The unproblematic cases in a higher place were done without the use of the equation. Yet the equation fits both the uncomplicated cases where branch resistors have the same resistance values and the more than difficult cases where branch resistors accept different resistance values. For instance, consider the application of the equation to the i simple and one difficult example below.

Example 1: Three 12 Ω resistors are placed in parallel	1/R_eq = i/R₁ + ane/R_ii + 1/R₃ ane/R_eq = i/(12 Ω) + ane/(12 Ω) + ane/(12 Ω) Using a calculator ... 1/R_eq = 0.25 Ω^-1 R_eq = i / (0.25 Ω^-one) R_eq = four.0 Ω
Instance 2: A 5.0 Ω, 7.0 Ω, and 12 Ω resistor are placed in parallel	ane/R_eq = i/R_i + i/R₂ + 1/R₃ 1/R_eq = 1/(five.0 Ω) + 1/(vii.0 Ω) + one/(12 Ω) Using a calculator ... i/R_eq = 0.42619 Ω-1 R_eq = 1 / (0.42619 Ω^-i) R_eq = 2.iii Ω

It's Your Turn to Attempt It

Need more practice? Use the Two Resistors in Parallel widget below to attempt some additional problems. Enter any ii resistance values y'all wish. Use your reckoner to determine the values of R_eq. And so click the Submit button to bank check your answers. Try it every bit many times as you wish with different resistance values.

Voltage Drops for Parallel Branches

It has been emphasized throughout the Circuits unit of The Physics Classroom tutorial that any voltage boost is acquired by a charge in the battery is lost by the accuse equally it passes through the resistors of the external excursion. The full voltage drop in the external excursion is equal to the gain in voltage equally a accuse passes through the internal circuit. In a parallel excursion, a charge does not pass through every resistor; rather, information technology passes through a single resistor. Thus, the entire voltage drop beyond that resistor must match the bombardment voltage. Information technology matters not whether the charge passes through resistor 1, resistor 2, or resistor 3, the voltage drib across the resistor that it chooses to laissez passer through must equal the voltage of the bombardment. Put in equation class, this principle would be expressed every bit

V_battery = V₁ = V₂ = V₃ = ...

If three resistors are placed in parallel branches and powered by a 12-volt battery, so the voltage drib beyond each one of the three resistors is 12 volts. A charge flowing through the circuit would only come across one of these 3 resistors and thus encounter a single voltage drop of 12 volts.

Electric potential diagrams were introduced in Lesson i of this unit and later used to illustrate the sequent voltage drops occurring in serial circuits. An electric potential diagram is a conceptual tool for representing the electric potential difference between several points on an electrical excursion. Consider the circuit diagram below and its corresponding electrical potential diagram.

Every bit shown in the electrical potential diagram, positions A, B, C, East and One thousand are all at a high electric potential. A single accuse chooses simply one of the three possible pathways; thus at position B, a single charge will move towards bespeak C, E or Grand and and so passes through the resistor that is in that branch. The accuse does not lose its high potential until information technology passes through the resistor, either from C to D, Due east to F, or G to H. One time information technology passes through a resistor, the accuse has returned to nearly 0 Volts and returns to the negative final of the bombardment to obtain its voltage boost. Unlike in series circuits, a accuse in a parallel circuit encounters a unmarried voltage driblet during its path through the external circuit.

The current through a given co-operative can be predicted using the Ohm's law equation and the voltage drop across the resistor and the resistance of the resistor. Since the voltage drop is the aforementioned across each resistor, the gene that determines that resistor has the greatest current is the resistance. The resistor with the greatest resistance experiences the lowest current and the resistor with the least resistance experiences the greatest electric current. In this sense, it could be said that charge (like people) chooses the path of least resistance. In equation class, this could be stated equally

I₁ = Δ Five₁ / R₁

I₂ = Δ 5₂ / R₂

I_iii = Δ V_three / R_three

This principle is illustrated by the circuit shown below. The product of I•R is the same for each resistor (and equal to the battery voltage). Yet the current is different in each resistor. The current is greatest where the resistance is least and the electric current is least where the resistance is greatest.

Mathematical Analysis of Parallel Circuits

The above principles and formulae can be used to analyze a parallel excursion and decide the values of the electric current at and electrical potential difference across each of the resistors in a parallel circuit. Their use will be demonstrated by the mathematical analysis of the excursion shown below. The goal is to use the formulae to decide the equivalent resistance of the circuit (R_eq), the current through the battery (I_tot), and the voltage drops and electric current for each of the iii resistors.

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

1 / R_eq = 1 / R_i + 1 / R₂ + 1 / R₃ = (1 / 17 Ω) + (ane / 12 Ω) + (i / 11 Ω)

1 / R_eq = 0.23306 Ω^-1

R_eq = i / (0.23306 Ω^-one)

R_eq = 4.29 Ω

(rounded from 4.29063 Ω)

Now that the equivalent resistance is known, the current in the battery can exist determined using the Ohm'south police equation. In using the Ohm'south law equation (ΔV = I • R) to determine the electric current in the battery, it is of import to employ the battery voltage for ΔV and the equivalent resistance for R. The adding is shown here:

I_tot = ΔV_battery / R_eq = (60 V) / (4.29063 Ω)

I_tot = 14.0 amp

(rounded from xiii.98396 amp)

The threescore Five battery voltage represents the gain in electric potential by a charge as it passes through the bombardment. The accuse loses this same amount of electric potential for any given laissez passer through the external circuit. That is, the voltage driblet across each 1 of the three resistors is the same as the voltage gained in the battery:

ΔV _battery = ΔV₁ = ΔV₂ = ΔV_three = 60 V

There are iii values left to be determined - the current in each of the individual resistors. Ohm's police is used once more to decide the current values for each resistor - it is simply the voltage drib across each resistor (60 Volts) divided by the resistance of each resistor (given in the problem statement). The calculations are shown below.

I₁ = ΔV₁ / R_i

I_i = (60 Five) / (17 Ω)

I₁ = 3.53 amp

I_ii= ΔV ₂ / R₂

I₂ = (60 V) / (12 Ω)

I₂ = 5.00 amp

I₃= ΔV _iii / R₃

I_iii = (60 5) / (11 Ω)

I_iii = 5.45 amp

As a bank check of the accuracy of the mathematics performed, it is wise to run across if the calculated values satisfy the principle that the sum of the current values for each individual resistor is equal to the total current in the circuit (or in the battery). In other words, is I_tot = I_ane + I₂ + I₃ ?

Is I_tot = I₁ + I_ii + I_three ?

Is fourteen.0 amp = iii.53 amp + 5.00 amp + 5.45 amp ?

Is 14.0 amp = 13.98 amp ?

Yes!!

(The 0.02 amp deviation is just the issue of having previously rounded the I_tot value from xiii.98.)

The mathematical assay of this parallel circuit involved a blend of concepts and equations. Equally is oft the case in physics, the divorcing of concepts from equations when embarking on the solution to a physics problem is a unsafe act. Hither, ane must consider the concepts that the voltage drops across each one of the 3 resistors is equal to the battery voltage and that the sum of the current in each resistor is equal to the total current. These understandings are essential in order to consummate the mathematical analysis. In the adjacent office of Lesson 4, combination or compound circuits in which some devices are in parallel and others are in series will exist investigated.

More Exercise

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

Nosotros Would Like to Suggest ...

Why but read about information technology and when you could be interacting with it? Interact - that's exactly what you do when you use one of The Physics Classroom'south Interactives. We would like to advise that you combine the reading of this folio with the use of our DC Excursion Builder Interactive. You tin observe it in the Physics Interactives section of our website. The DC Circuit Builder provides the learner with a virtual circuit building kit. You can easily drag voltage sources, resistors and wires onto the workspace and arrange and connect them anyway you wish. Voltmeters and ammeters allow you lot to measure current and voltage drops. Tapping a resistor or a voltage source allows you to change the resistance or the input voltage. It's easy. It's fun. And it's safe (unless yous're using information technology in the bathtub).

Check Your Understanding

1. Equally more and more resistors are added in parallel to a circuit, the equivalent resistance of the circuit ____________ (increases, decreases) and the total current of the excursion ____________ (increases, decreases).

ii. Iii identical low-cal bulbs are connected to a D-cell as shown below. P, Q, X, Y and Z stand for locations forth the circuit. Which i of the following statements is true?

a. The current at Y is greater than the current at Q.
b. The current at Y is greater than the electric current at P.

c. The electric current at Y is greater than the current at Z.

d. The electric current at P is greater than the current at Q.

e. The electric current at Q is greater than the electric current at P.

f. The current is the same at all locations.

3. Three identical lite bulbs are connected to a D-cell as shown below. P, Q, X, Y and Z represent locations along the circuit. At which location(s), if any, will the current exist ...

a. ... the same as at Ten?
b. ... the same every bit at Q?

c. ... the same as at Y?

d. ... less than at Q?

east. ... less than at P?

f. ... twice that at Z?

k. ... three times that at Y?

4. Which adjustments could be made to the circuit beneath that would decrease the current in the cell? List all that apply.

a. Increase the resistance of bulb X.
b. Subtract the resistance of seedling X.

c. Increase the resistance of bulb Z.

d. Decrease the resistance of seedling Z.

e. Increase the voltage of the cell (somehow).

f. Subtract the voltage of the jail cell (somehow).

g. Remove bulb Y.

5. A 12-V battery , a 12-ohm resistor and a 4-ohm resistor are continued as shown. The current in the 12-ohm resistor is ____ that in the 4-ohm resistor.

a. i/3

b. 1/ii

c. 2/three

d. the aforementioned as

e. ane.v times

f. twice

g. three times

h. 4 times

6. A 12-Five bombardment , a 12-ohm resistor and a 4-ohm resistor are connected as shown. The voltage drop across the 12-ohm resistor is ____ that beyond the 4-ohm resistor.

a. one/3

b. 1/two

c. 2/iii

d. the same equally

e. ane.five times

f. twice

thousand. 3 times

h. four times

seven. A 12-Five battery and a 12-ohm resistor are connected as shown in circuit. A 6-ohm resistor is added to the 12-ohm resistor to create circuit Y as shown. The voltage drop across the 6-ohm resistor in circuit Y is ____ that across the resistor in X.

a. larger than
b. smaller than

c. the same every bit

8. Use your understanding of equivalent resistance to complete the following statements:

a. Two vi-Ω resistors placed in parallel would provide a resistance that is equivalent to one _____-Ω resistor.
b. Three six-Ω resistors placed in parallel would provide a resistance that is equivalent to one _____-Ω resistor.

c. Iii eight-Ω resistors placed in parallel would provide a resistance that is equivalent to i _____-Ω resistor.

d. Three resistors with resistance values of 2-Ω, 4-Ω, and 6-Ω are placed in parallel. These would provide a resistance that is equivalent to ane _____-Ω resistor.

due east. Iii resistors with resistance values of 5-Ω, 6-Ω, and 7-Ω are placed in parallel. These would provide a resistance that is equivalent to 1 _____-Ω resistor.

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

nine. Based on your answers to the higher up question, consummate the following argument:

The overall or equivalent resistance of three resistors placed in parallel volition be _____.

a. greater than the resistance of the biggest R value of the three.
b. less than the resistance of the smallest R value of the three.

c. somewhere in between the smallest R and the biggest R value of the three.

d. ... nonsense! No such generalization tin can be made. The results vary.

ten. Three resistors are continued in parallel. If placed in a excursion with a 12-volt power supply. Determine the equivalent resistance, the total excursion current, and the voltage drop across and current in each resistor.