How do two resistors in parallel, one 30 ohms, the other 20 ohms, reduce to a resister of 12 ohms in series?

Answer 1

Do the calculations for the parallel arrangement and you get the answer

Answer 2

They do this following way. Two resistors in parallel have the same voltage on their ends. But the sum of the currents flowing through each of them equals to the current which flows over a resistor of 12 Ohm by the same voltage.
For instance, if the voltage is 30 V over the first resistor flows 1 A and over the second - 1,5 A. (Voltage/Resistor value) The sum of currents is 2,5 A. 30 V divided by 2,5 A give us exactly 12 Ohm.

By the series connection two resistors have the same current but the voltage drops on each of them are in the ratio as a proportion of their values.

Answer 3

the series-parallel combination in resistance law in electric states that when the two resistance are in parallel they get multiplied and when the two resistance are in series they get added. R1 x R2 = 30 x 20 = 600, R1R2 =30+20=50
600 divide 50 equals to 12 ohms.

Answer 4

because a 30 ohm resistor has a conductance of 1/30 = .0333 mhos (or siemens), a 20 ohms resistor has a conductance of 1/20 =.05 mhos, and since conductances arranged in parallel add, the total conductance is .0333+ 0.05 = .08333 mhos, which is equivalent to a resistance of 1/.08333 = 12 ohms

Answer 5

Because electrons take the path of least resistance in proportion to the resistance of each parallel resistor.

Answer 6

Well, the formula is R1timesR2 divided R1 plus R2 but I couldn't begin to prove that formula... saw it done once by solving for R1 and it took up every blackboard around the perimeter of the classroom

Answer 7

1/30+1/20=83.333m

1/83.3333m=12 ohms

its called the reciprical method

Answer 8

Rt = R1 X R2 DIVIDED BY R1 + R2

Answer 9

1/r=1/r'+i/r''+1/r'''+.....so on
for your problem-
1/r=1/30+1/20
1/r=50/30*20
r=30*20/50
r=12ohms
thats your answers.....