Determine the current and power dissipated in the resistors
Image from: Irwin, J. David., and R. M. Nelms. Basic Engineering Circuit Analysis, Tenth Edition. N.p.: John Wiley & Sons, 2010. Print.
Solution:
Looking at the diagram, we see that one of the elements is represented in Siemens value (the 0.5 S – which represents conductance). We will first convert this into ohms. Remember that conductance, G, is equal to \frac{1}{R} where R is resistance.
Thus, we can write:
0.5=\frac{1}{R}
R=2\Omega
Now that we know the resistance value, we can see that all the resistors are in series, meaning you can add them up together, which gives us a total resistance value of 4\Omega.
To figure out the current, remember that current,I, is equal to voltage divided by the resistance.
I=\frac{V}{R}
Substituting the values we have gives is:
I=\frac{12}{4}
I=3A
Let us now figure out the power dissipated by each resistor.
Remember that power, P=(I^2)(R)
For the both 2\Omega resistors:
P=(3^2)(2)
P=18W
This question can be found in Basic Engineering Circuit Analysis, 10th edition, chapter 1, question 2.2
