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.
