When a 10 ohm resistor is connected in parallel to a 15 ohm resistor what percentage of the total current will flow through the 10 ohm resistor?

Answer 1

There will be 20 amps flowing through the 5 ohm resistor. We could do some math and all to figure voltage drops and some other things, but let's cut to the chase and see what's happening. Based on what we know about series circuits and about parallel circuits, we can shred this in nothing flat. So let's. First, we're told 30 amps flows in the circuit. That's the total current, and it will be the current through the first 10 ohm resistor. It has to be. The the 30 amps "splits" to flow through the parallel network of the 10 ohm and 5 ohm resistors. That's 30 amps that has to "split" and some will go through the 10 ohm resistor and some will go through the 5 ohm resistor. With me? Sure. Now for the "trick" here. Since the 5 ohm resistor has only half the resistance of the 10 ohm resistor, twice as much current will flow through it as the 10 ohm resistor. Make sense? Yup. Let's finish this. Since there is twice as much current flowing in the 5 ohm resistor 'cause it's got half the resistance of the 10 ohm resistor, if we have "x" amount of current flowing in the 10 ohm resistor, then we'll have "2x" amps of current flowing in the 5 ohm resistor. That's "3x" amps total, and the "3x" amps equals 30 amps. See through it now? There will be 10 amps flowing through the 10 ohm resistor, and 20 amps flowing through the 5 ohm resistor. Piece of cake.

Answer 2

The current in a passive DC parallel circuit divides in inverse proportion to the
resistance of the branch, as I recall. So, since the 10-ohm is 2/3 the resistance
of the 15-ohm, it ought to carry 3/2 as much current. We can call the current
through the 10-ohm "3/2" and the current through the 15-ohm "1". We don't care
what they are in actual Amps, but the ratio is correct. Then the total current in
both branches is 2.5, and the 10-ohm is carrying 1.5, so that's 60% of the total.


Now, I hafta tellya . . . I was really winging it there, engineering by the seat of
my pants as it were, and I have little confidence in the result. So before we part,
I must do it the old-fashioned way as a check:

-- (10 ohms) in parallel with (15 ohms) = 10 x 15/10 + 15 = 150/25 = 6 ohms effective.

-- Connect a 6-volt battery across the parallel resistors. Total current = 6/6 = 1 Amp.

-- Current through the 10-ohm resistor = E/R = 6/10 = 0.6 Amp = 60% of the total.

Yay !