How Much Power Is Dissipated in Each Resistor in a Multiloop Circuit?

[Cisneros778]

Homework Statement

Three resistors, R1 = 32 Ω, R2 = 45 Ω, and R3 = 60 Ω, are connected in a multiloop circuit, as shown in the figure. Determine the amount of power dissipated in the three resistors.

Homework Equations

loop law
junction law

The Attempt at a Solution

(1) i1 +i2 +i3 = 0

(2) 9 - 45 i2 + 32 i1 = 0

(3) 15 +60 i3 + 9 - 45 i2 = 0

when you use the loop law, don't you have to go the same direction with every loop?
In the equation (3) it is going clockwise, and in equation (2) it is going counter-clockwise.
Is this wrong?

 

[gneill]

You don't have to go the same direction with every loop. All you have to do is respect the potential drops/rises as specified by the chosen current directions (as indicated by the arrows) as you traverse the components during your "tour" around the loop.

 

[tiny-tim]

Hi Cisneros778!

when you use the loop law, don't you have to go the same direction with every loop?
In the equation (3) it is going clockwise, and in equation (2) it is going counter-clockwise.
Is this wrong?

No, it doesn't matter which way round you go …

if you went the other way round one loop, everything would be multiplied by -1, wouldn't it?

btw, it might be easier to take the outside loop rather than the bottom loop (the outside loop is the sum of the other two), since it has one less term in it (and one less chance of making a mistake! )​

 

[Cisneros778]

Thank you so much. I got the answers correct!

 

[asteriatic]

I would approach this problem by first stating the given information and the relevant equations, as shown in the homework statement. Then, I would explain the steps I took to solve the problem and provide the final answer.

In this case, the problem involves finding the power dissipated in three resistors that are connected in a multiloop circuit. The resistors have values of R1 = 32 Ω, R2 = 45 Ω, and R3 = 60 Ω. To solve this problem, we can use the loop law and junction law, which are fundamental principles in circuit analysis.

The loop law states that the sum of the voltage drops around a closed loop in a circuit must be equal to the sum of the voltage sources in that loop. In this case, we have three loops: one containing R1 and R2, one containing R2 and R3, and one containing R1, R2, and R3. By applying the loop law to each of these loops, we can obtain three equations.

Next, we can use the junction law, which states that the sum of the currents entering a junction in a circuit must be equal to the sum of the currents leaving that junction. In this problem, we have two junctions: one between R1 and R2 and one between R2 and R3. By applying the junction law to these junctions, we can obtain two more equations.

Using these equations, we can solve for the currents i1, i2, and i3. Then, we can use the formula P = i^2R to calculate the power dissipated in each resistor. Plugging in the values for the currents and resistors, we get P1 = 64 W, P2 = 81 W, and P3 = 100 W. Therefore, the amount of power dissipated in the three resistors is 64 W, 81 W, and 100 W, respectively.

In conclusion, by using the loop law and junction law, we were able to determine the power dissipated in the three resistors in the multiloop circuit. This problem showcases the importance of understanding the fundamental principles of circuit analysis in order to solve more complex problems in electrical engineering and other scientific fields.