# Circuit analysis

• Engineering
i dont need help writing the c++ code, but i do need help calculating the power through the resistor. P=VI and P = I^2R aren't working, so I'm guessing it's got something to do with circuit analysis, which i haven't been taught yet, and that my c++ class hasn't gone over.

------2 ohms---- ----3 ohms----
| | | |
| | | |
--- ---- -------
| | | | | |
| | | | | |
| ------4 ohms---- -----6 ohms---- |
| |
100 volts 10 - 20 ohms
| |
| |
___ ___
- -

In the above circuit there is a 100 volt DC source, a 2 ohm
resistor in parallel with a 4 ohm resistor, and then those
resistors are in series with a 3 ohm resistor in parallel
with a 6 ohm resistor. The circuit is terminated in a load
whose value varies randomly between 10 and 20 ohms.

If the power dissipated in the 2 ohm resistor goes above 35
watts, the resistor will burn and its resistance will jump
from 2 ohms to 1000000 ohms.

Solve the circuit 100,000 times (using random values for the
load). During the 100,000 simulations, count the number of times that
the 2 ohm resistor burns up. Also, calculate the average
voltage delivered to the load resistor when the 2 ohm
resistor has burned up.

Your program should output only two numbers on separate lines
as follows:

1-Number of times that the 2 ohm resistor burns

2-Average voltage delivered to the load resistor calculated
over the cases where the 2 ohm resistor has burned up

The seed to use for the random number generator is 100000

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Integral
Staff Emeritus
Gold Member
You need to tell us what expression you are using for the power through the 2 ohm resistor. That is exactly how are you calculating it. What is the current? How did you arrive at it?

Just some hints:

Find the current flowing through the load resistor by determining the total resistance of the circuit and applying ohms law.

You can find the resistance of two resistors in parallel by using (R1 * R2) / (R1 + R2).

You can find the current flowing through the 2 ohm resistor by using I / 3 * 2, where I is the current flowing through the load resistor.