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

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This question can be found in Basic Engineering Circuit Analysis, 10th edition, chapter 1, question 2.2