Parallel circuits Watch

K-pop02
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What happens to the current in a parallel circuit when it reaches a branch in the circuit?
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CoreyTheMans
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It splits half carries on and the other half travels the branch
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K-pop02
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Thank You
(Original post by CoreyTheMans)
It splits half carries on and the other half travels the branch
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Alex-1234
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Hello
I am a qualified Electronics Engineer and I will try and explain this phenomenon to you

If you have, say Two resistors in Parallel, values R1 = 10 ohms and R2 = 25 ohms you have a TOTAL circuit Resistance of R1R2/R1 + R2 which is approx 10 x 25 / 35 = 7.1 ohms
The circuit voltage is 21 volts. This means by OHMS LAW that the circuit current is Circuit Voltage / Total circuit resistance = 21 / 7.1 = 2.96 Amps

Right, clear so far? This 2.96 A current will start to flow around the network but when it hits the PARALLEL bit it has to split in TWO. The question is how much current will flow through the 10-ohm resistor and how much goes through the 25-ohm resistor?

Well you should know that the Voltage across both resistors is the SAME as the circuit voltage i.e. 21 volts

So the current flowing in the 10 - ohm resistor is = 21 / 10 = 2.1 Amps
Likewise, the current in the 25-ohm resistor is = 21/25 = 0.85 Amps

Now when the current reappears again from the parallel network it should regroup and be the same value it was BEFORE it entered the parallel circuit YES? So we can check by adding the two circuit leg currents and we should arrive back at the original value of 2.96A

So ADD Leg 1 current which is 2.1 A to the Leg 2 current of 0.85 A and we get 2,96 A ........the original figure we calculated earlier.

Its relatively easy when you only have TWO LEGS try tackling a 4 or 5 leg series-parallel network and it can get quite tricky.

Good Luck
(Original post by K-pop02)
What happens to the current in a parallel circuit when it reaches a branch in the circuit?
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K-pop02
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Thank you so much for this explanation
(Original post by Alex-1234)
Hello
I am a qualified Electronics Engineer and I will try and explain this phenomenon to you

If you have, say Two resistors in Parallel, values R1 = 10 ohms and R2 = 25 ohms you have a TOTAL circuit Resistance of R1R2/R1 + R2 which is approx 10 x 25 / 35 = 7.1 ohms
The circuit voltage is 21 volts. This means by OHMS LAW that the circuit current is Circuit Voltage / Total circuit resistance = 21 / 7.1 = 2.96 Amps

Right, clear so far? This 2.96 A current will start to flow around the network but when it hits the PARALLEL bit it has to split in TWO. The question is how much current will flow through the 10-ohm resistor and how much goes through the 25-ohm resistor?

Well you should know that the Voltage across both resistors is the SAME as the circuit voltage i.e. 21 volts

So the current flowing in the 10 - ohm resistor is = 21 / 10 = 2.1 Amps
Likewise, the current in the 25-ohm resistor is = 21/25 = 0.85 Amps

Now when the current reappears again from the parallel network it should regroup and be the same value it was BEFORE it entered the parallel circuit YES? So we can check by adding the two circuit leg currents and we should arrive back at the original value of 2.96A

So ADD Leg 1 current which is 2.1 A to the Leg 2 current of 0.85 A and we get 2,96 A ........the original figure we calculated earlier.

Its relatively easy when you only have TWO LEGS try tackling a 4 or 5 leg series-parallel network and it can get quite tricky.

Good Luck
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