# Need help with circuits question.

1. Feb 2, 2009

### crhscoog

1. The problem statement, all variables and given/known data

3. The attempt at a solution

i. First I combined 20ohm and 30ohm to make 50ohms. Since this 50ohms and 100ohms is in parallel I have to use 1/Rtotal= 1/r1 + 1/r2.... etc and got 33.333 ohms. I added this to 10ohms which ends up being 43.333 ohms. Then I used V=IR and got I=.277. So would .277 be the current going through 10ohm and half of that going through 100ohm and quarter of .277 going through 20ohm and 30ohm???

So from here I would get half of .277 as current for 100ohm and a fourth of .277 as current for both 20 and 30 ohm. and using P=RI^2 i would get a total of 2.925W for power consumption??

and with 2.925 I determine that it takes 3418.803419 hours to provide 10kJ of electrical energy??

Last edited: Feb 2, 2009
2. Feb 3, 2009

### brasidas

1. $I = 0.277A$ right? Total power consumption assumes the total resistance, which is $R = 43.3\Omega$, right? So when I carried out the calculation, I ended up with $P = 3.322 W$

2. $P = 3.322 W = 3.322 \frac{J}{s}$ Therefore, $10000 J = 3.322\frac{J}{s} \cdot x(s)$, right? So you will end up with $x = \frac{10000 J}{3.322\frac{J}{s}} = 3010(s)$

$$\frac{3010}{3600}h = 0.836 h = 50.16 min$$

So 50 minutes and 10 seconds is what it takes to provide 10kJ of energy.