Showing posts with label #resistors in parallel. Show all posts
Showing posts with label #resistors in parallel. Show all posts

Monday 21 June 2021

Resistors in series and Parallel

 Resistor

  • limits the flow of charge in a circuit and is an ohmic device where V=IR.
  • Most circuits have more than one resistor.
  • If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit.

Resistor in series

 

  • Resistors are said to be in series whenever the current flows through the resistors sequentially.
  • The Figure which shows three resistors in series with an applied voltage equal to Vab.
  • Since there is only one path for the charges to flow through, the current is the same through each resistor.
  • The equivalent resistance of a set of resistors in a series connection is equal to the algebraic sum of the individual resistances.

Example 1

A battery with a terminal voltage of 9V is connected to a circuit consisting of four 20Ω and one 10Ω resistors all in series Assume the battery has negligible internal resistance.

(a)    Calculate the equivalent resistance of the circuit.

(b)    Calculate the current through each resistor.

(c)     Calculate the potential drop across each resistor.

(d)   Determine the total power dissipated by the resistors and the power supplied by the battery.

Solution 1

  1. The equivalent resistance is the algebraic sum of the resistances
  2. The current through the circuit is the same for each resistor in a series circuit   and is equal to the applied voltage divided by the equivalent resistance:
  3. The potential drop across each resistor can be found using Ohm’s law:

 


Note that the sum of the potential drops across each resistor is equal to the voltage supplied by the battery


Resistors in Parallel

  • Resistors are in parallel when one end of all the resistors are connected by a continuous wire of negligible resistance and the other end of all the resistors are also connected to one another through a continuous wire of negligible resistance.
  • The potential drop across each resistor is the same.
  • Current through each resistor can be found using Ohm’s law I=V/R, where the voltage is constant across each resistor.
  • For example, an automobile’s headlights, radio, and other systems are wired in parallel, so that each subsystem utilizes the full voltage of the source and can operate completely independently. The same is true of the wiring in your house or any building.
  • The junction rule gives I=I1+I2.
  • There are two loops in this circuit, which leads to the equations V=I1Rand I1R1=I2R2
  • Note the voltage across the resistors in parallel are the same (V=V1=V2) and the current is additive:

Example 2

Three resistors R_1=1.00Ω, R_2=2.00Ω, and R_3=2.00Ω are connected in parallel. The parallel connection is attached to a V=3.00V 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.

Solution

  1. The total resistance for a parallel combination of resistors is found using Equation. Entering known values gives


  1. The total current can be found from Ohm’s law, substituting Req for the total resistance. This gives

Current I for each device is much larger than for the same devices connected in series

The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus,

 


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=V2/R, since each resistor gets full voltage. Thus,


e. The total power can also be calculated in several ways. Choosing  P=IV and entering the total current yields