Given the circuit, find the voltage across each resistor and the power dissipated in each.

Image from: Irwin, J. David., and R. M. Nelms. Basic Engineering Circuit Analysis, Tenth Edition. N.p.: John Wiley & Sons, 2010.

#### Solution:

### Let us first convert 0.25 S into ohms. Currently, the value is in Siemens, representing conductance.

### 0.25 = \frac{1}{R}

### R=4\Omega

### Remember that voltage across a resistor can be found using V=IR, where V is voltage, I is current and R is resistance.

### For the 5\Omega resistor, we have the following:

### V=(6A)(5\Omega)

### V=30V

### For the 4\Omega resistor, we have the following:

### V=(6A)(4\Omega)

### V=24V

### To find the power dissipated, we can use the following formula:

### P=\frac{V^2}{R} where P is power represented in watts.

### For our 5\Omega resistor, we have:

### P=\frac{30^2}{5}

### P=180W

### For our 4\Omega resistor, we have:

### P=\frac{24^2}{4}

### P=144W

###### This question can be found in Basic Engineering Circuit Analysis, 10th edition, chapter 1, question 2.4.