In a parallel circuit,what happens to the current?

Question

Accurate answers please! Thanks!

Answer 1

It gets split up between each row depending on it's resistance, for example if there are 5 rows of wiring and 100 amperes (amps) the amount of amps supplied to each row would be:
100amps/5rows
= 20 amps per row (if the resistance was the same, which is not likely but it gives you an idea).
Try using GCSE/KS3 bitesize which would help you:
http://www.bbc.co.uk/schools/ks3bitesize/science/physics/electricity_4.shtml

Good luck!
Lorna

PS. If you get stuck, I suggest you ask your teacher because if you don't understand the basics, you will struggle later on!

Answer 2

Current sub - divides through the different 'branches' of circuit


Kirchoff's Current Law

At every node, the sum of all currents entering a node must equal zero. What this law means physically is that charge cannot accumulate in a node; what goes in must come out. In the example, Figure 1, below we have a three-node circuit and thus have three KCL equations.
−i−i1=0 (1)
i1−i2=0 (2)
i+i2=0 (3)
Note that the current entering a node is the negative of the current leaving the node.

Given any two of these KCL equations, we can find the other by adding or subtracting them. Thus, one of them is redundant and, in mathematical terms, we can discard any one of them. The convention is to discard the equation for the (unlabeled) node at the bottom of the circuit.

Answer 3

In a circuit containing a parallel-plate capacitor and a potential source (mutually with a battery), the potential source aspects potential for costs to circulate to the plates of the capacitor - unfavorable costs on one plate and effective ones on the different plate. This effectively creates an electric powered field between the plates of the capacitor. This field shops potential and it is the potential that flows around the area between the plates - not the fees. The circulate of potential between the plates effectively keeps the continuity of 'modern-day' between the capacior plates.

Answer 4

The current gets divided into each branch in an inverse proportion to the resistance of each branch. If R1, R2 and R3 are the resistances of the three branches (as an example), R is given by

1/R = 1/R1 + 1/R2 + 1/R3

Current I = V/R = V/R1 + V/R2 + V/R3

Answer 5

The current is shared in the branches in an inversely proportional fashion to the resistance

Answer 6

If the resistance of each of the parallel wires is the same, then the current will split equally/ 'half'.

Answer 7

Some goes through each branch in inverse ratio to the resistance of the branch.

Answer 8

The current is shared in inverse proportion to the impedance of the respective paths.