Resistors in a parallel circuit?
Three resistors (14.0 Ω, 18.9 Ω, 5.9 Ω) are connected in parallel across 107 V. What is the total resistance for this circuit?
this one is just: 1/14 + 1/18.9 +1/5.9 = 3.46 ohms
But these are what I don't know how to do....
What is the voltage drop across the 18.9 Ω resistor?
What current is flowing in the 5.9 Ω resistor?
What is the total circuit current?
What power is dissipated at the 14.0 resistor?
Update:I just figured out the current for 5.9, it's: 18.14
And the total current is: 31.44
Can't get the other one though.
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Verified answer
Resistors in parallel all receive the same voltage (they all have a potential difference equal to the source), so they will all dissipate the same amount of voltage here. Current through each is just the voltage divided by the resistor you're interested in. Total current can be found in two ways: take the voltage over the net resistance, or find the current through each of the three resistors and sum them up. It should work out the same, so it's a good way to check. Compare:
Inet = I1 + I2 + I3 = 107 V * (1/14 + 1/18.9 +1/5.9) = 31.4398 A
to:
I = V/Rnet = 107/3.4033 = 31.4398 A
Power dissipated can be found using V²/R or I²R, where I is the current only through that resistor. Either way works, though using voltage would be easier since both values are given.
EDIT: Sorry, read 107 as 170. Either way, if you get the principle, it's pretty simple.
EDIT2: What's not to get? The equation is V²/R and both are explicitly given. Spoilers, it turns out to 817.7857 W. If you want a simple check on this too, the sum of power absorbed by the resistors should equal power provided by the source.
P1 + P2 + P3 = ΣV²/R = 107² * (1/14 + 1/18.9 +1/5.9) = 3364.1 W
PS = I*V = (31.4398 A)(107 V) = 3364.1 W