Verified answer

1. Add the 12 Ohm resistor directly across the 30 Volt battery to produce the largest increase in total current.

Rt = [(4 ll 4)Ohm] + 12 Ohm + [ (6 + 3) ll 18] Ohm = (2 + 12 + 6)Ohm = 20 Ohms

it = 30V/20 Ohm = 1.5A

2. Power dissipated in the two 4 Ohm resistors = [(1.5A)^2]*(2 Ohm) = 4.5 Watts

Do the second part first. Get the total R. The two 4's equal 2, the 3+6 = 9 and that in parallel with 18 is 18*9/(9+18) = 6.

so total R = 2 + 12 + 6 = 20 ohms.

Total current is 30/20 = 1.5 amps

That splits evenly in the two 4's for 0.75 each, and the power in each is I²R = 2.25 watts each.

As to where you add the 12, putting it in series would decrease the current, so you have to put it in parallel with something. And to have the most effect would be in parallel with the largest R, which is the other 12.

enable's make it ordinary. we've a resistor: 10 ohms And yet another resistor: 20 ohms we've a voltage: 12 volts series resistance upload: entire = 30 ohms How lots modern is going by those resistors? I=E/R voltage / resistance 12 / 30 = 0.4 amps How lots is flowing by R2? the comparable quantity ! modern circulate in a chain circuit is the comparable at any factor. How lots potential is disipated via R2? potential is watts. Use Ohm's regulation back. P = I x E 0.4A x 12V = 4.8W It greater perfect be a huge resistor and it is probable warm.

You need to combine the resistor by using parallel and series combinations to get a single resistance. Once you do that you can find the current; call it I. The two 4 ohm resistors in parallel is 2 ohms. The total power consumed by them is I^2*R.