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R eq = ( 1 R 1 + 1 R 2 + 1 R 3 + + 1 R N 1 + 1 R N ) −1 = ( i = 1 N 1 R i ) −1 .

This relationship results in an equivalent resistance R eq that is less than the smallest of the individual resistances. When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, so the total resistance is lower.

Analysis of a parallel circuit

Three resistors R 1 = 1.00 Ω , R 2 = 2.00 Ω , and R 3 = 2.00 Ω , are connected in parallel. The parallel connection is attached to a V = 3.00 V voltage source. (a) What is the equivalent resistance? (b) Find the current supplied by the source to the parallel circuit. (c) Calculate the currents in each resistor and show that these add together to equal the current output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source and show that it equals the total power dissipated by the resistors.

Strategy

(a) The total resistance for a parallel combination of resistors is found using R eq = ( i 1 R i ) −1 .

(Note that in these calculations, each intermediate answer is shown with an extra digit.)

(b) The current supplied by the source can be found from Ohm’s law, substituting R eq for the total resistance I = V R eq .

(c) The individual currents are easily calculated from Ohm’s law ( I i = V i R i ) , since each resistor gets the full voltage. The total current is the sum of the individual currents: I = i I i .

(d) The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use P i = V 2 / R i , since each resistor gets full voltage.

(e) The total power can also be calculated in several ways, use P = I V .

Solution

  1. The total resistance for a parallel combination of resistors is found using [link] . Entering known values gives
    R eq = ( 1 R 1 + 1 R 2 + 1 R 3 ) −1 = ( 1 1.00 Ω + 1 2.00 Ω + 1 2.00 Ω ) −1 = 0.50 Ω .

    The total resistance with the correct number of significant digits is R eq = 0.50 Ω . As predicted, R eq is less than the smallest individual resistance.
  2. The total current can be found from Ohm’s law, substituting R eq for the total resistance. This gives
    I = V R eq = 3.00 V 0.50 Ω = 6.00 A .

    Current I for each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallel connections has a smaller total resistance than the resistors connected in series.
  3. The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus,
    I 1 = V R 1 = 3.00 V 1.00 Ω = 3.00 A .

    Similarly,
    I 2 = V R 2 = 3.00 V 2.00 Ω = 1.50 A

    and
    I 3 = V R 3 = 6.00 V 2.00 Ω = 1.50 A .

    The total current is the sum of the individual currents:
    I 1 + I 2 + I 3 = 6.00 A .
  4. The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use P = V 2 / R , since each resistor gets full voltage. Thus,
    P 1 = V 2 R 1 = ( 3.00 V ) 2 1.00 Ω = 9.00 W .

    Similarly,
    P 2 = V 2 R 2 = ( 3.00 V ) 2 2.00 Ω = 4.50 W

    and
    P 3 = V 2 R 3 = ( 3.00 V ) 2 2.00 Ω = 4.50 W .
  5. The total power can also be calculated in several ways. Choosing P = I V and entering the total current yields
    P = I V = ( 6.00 A ) ( 3.00 V ) = 18.00 W .

Questions & Answers

instrument for measuring highest temperature of a body is?
brian Reply
Thermometer
Umar
how does beryllium decay occur
Sandy Reply
Photon?
Umar
state the first law of thermodynamics
Kansiime Reply
Its state that "energy can neither be created nor destroyed but can be transformed from one form to another. "
Ayodamola
what about the other laws can anyone here help with it please
Sandy
The second law of thermodynamics states that the entropy of any isolated system always increases. The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero.
sahil
The first law is very simple to understand by its equation. The law states that "total energy in thermodynamic sytem is always constant" i.e d¶=du+dw where d¶=total heat du=internal energy dw=workdone... PLEASE REFER TO THE BOOKS FOR MORE UNDERSTANDING OF THE CONCEPT.
Elia
what is distance.?
Ali Reply
what is physics?
Ali
Physics is a scientific phenomenon that deals with matter and its properties
Ayodamola
physics is the study of nature and science
John
Chater1to7
min
Physics is branch of science which deals with the study of matters in relation with energy.
Elia
What is differential form of Gauss's law?
Rohit Reply
help me out on this question the permittivity of diamond is 1.46*10^-10.( a)what is the dielectric of diamond (b) what its susceptibility
OLUWA Reply
a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
Abu Reply
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
Lord Reply
kya ye b.sc ka hai agar haa to konsa part
MPL Reply
what is charge quantization
Mayowa Reply
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
Aditya
determine absolute zero
OFERE Reply
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
Opeyemi Reply
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
Ashok Reply
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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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