Explain how three resistors can be connected in parallel
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Each resistor in the circuit has the full voltage. According to Ohm’s law, the currents flowing through the individual resistors are I1=VR1I1=VR1, I2=VR2I2=VR2, and I3=VR3I3=VR3. Conservation of charge implies that the total current is the sum of these currents:
Parallel resistors: Three resistors connected in parallel to a battery and the equivalent single or parallel resistance.
I=I1+I2+I3.I=I1+I2+I3.
Substituting the expressions for individual currents gives:
I=VR1+VR2+VR3I=VR1+VR2+VR3
or
I=V(1R1+1R2+1R3)I=V(1R1+1R2+1R3)
This implies that the total resistance in a parallel circuit is equal to the sum of the inverse of each individual resistances. Therefore, for every circuit with nn number or resistors connected in parallel,
Rn(parallel)=1R1+1R2+1R3…+1Rn.Rn(parallel)=1R1+1R2+1R3…+1Rn.
This relationship results in a total resistance 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.
Each resistor in parallel has the same full voltage of the source applied to it, but divide the total current amongst them. This is exemplified by connecting two light bulbs in a parallel circuit with a 1.5V battery. In a series circuit, the two light bulbs would be half as dim when connected to a single battery source. However, if the two light bulbs were connected in parallel, they would be equally as bright as if they were connected individually to the battery. Because the same full voltage is being applied to both light bulbs, the battery would also die more quickly, since it is essentially supplying full energy to both light bulbs. In a series circuit, the battery would last just as long as it would with a single light bulb, only the brightness is then divided amongst the bulbs.
Parallel resistors: Three resistors connected in parallel to a battery and the equivalent single or parallel resistance.
I=I1+I2+I3.I=I1+I2+I3.
Substituting the expressions for individual currents gives:
I=VR1+VR2+VR3I=VR1+VR2+VR3
or
I=V(1R1+1R2+1R3)I=V(1R1+1R2+1R3)
This implies that the total resistance in a parallel circuit is equal to the sum of the inverse of each individual resistances. Therefore, for every circuit with nn number or resistors connected in parallel,
Rn(parallel)=1R1+1R2+1R3…+1Rn.Rn(parallel)=1R1+1R2+1R3…+1Rn.
This relationship results in a total resistance 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.
Each resistor in parallel has the same full voltage of the source applied to it, but divide the total current amongst them. This is exemplified by connecting two light bulbs in a parallel circuit with a 1.5V battery. In a series circuit, the two light bulbs would be half as dim when connected to a single battery source. However, if the two light bulbs were connected in parallel, they would be equally as bright as if they were connected individually to the battery. Because the same full voltage is being applied to both light bulbs, the battery would also die more quickly, since it is essentially supplying full energy to both light bulbs. In a series circuit, the battery would last just as long as it would with a single light bulb, only the brightness is then divided amongst the bulbs.
Answered by
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Three resistors can be connected in parallel when the branches of current are different and they form like the attachment :
The current will divide in the branches .
Hence I = I1 + I2 + I3
= > V/R = V/R1 + V/R2 + V/R3
= > 1/R = 1/R1 + 1/R2 + 1/R3
Here was the derivation .
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