CIRCUIT ANALYSIS: Find the equivalent resistance seen by the source

In summary, using series/parallel resistance combination, the equivalent resistance seen by the source in the circuit is 40 Ohms. The overall dissipated power is 3.6W, calculated using the formula P = V^2/R. The process involved adding resistors in series and using the formula for resistors in parallel to find the equivalent resistance.
  • #1
VinnyCee
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Homework Statement



Using series/parallel resistance combination, find the equivalent resistance seen by the source in the circuit below. Find the overall dissipated power.

http://img300.imageshack.us/img300/6132/chapter2problem34au4.jpg [Broken]

Homework Equations



For resistors in series: [tex]R_{eq}\,=\,R_1\,+\,R_2\,+\,\cdots\,+\,R_n[/tex]

For resistors in parallel: [tex]R_{eq}\,=\,\frac{1}{R_1}\,+\,\frac{1}{R_2}\,+\,\cdots\,+\,\frac{1}{R_n}[/tex]

Also for resistors in parallel: [tex]R_{eq}\,=\,\frac{R_1\,R_2}{R_1\,+\,R_2}[/tex]

The Attempt at a Solution



Adding the right three resistors that are in series to get an equivalent one that is 40 Ohms.

http://img205.imageshack.us/img205/1186/chapter2problem34part2dk8.jpg [Broken]

Using the formula above for parallel resistors: [tex]R_{eq}\,=\,\frac{(40\Omega)\,(40\Omega)}{(40\Omega)\,+\,(40\Omega)}\,=\, 20\Omega[/tex]

http://img391.imageshack.us/img391/7092/chapter2problem34part3ur9.jpg [Broken]

Again, combining the resistors on the right that are in series.

http://img205.imageshack.us/img205/5549/chapter2problem34part4jn1.jpg [Broken]

Using the parallel formula again: [tex]R_{eq}\,=\,\frac{(40\Omega)\,(40\Omega)}{(40\Omega)\,+\,(40\Omega)}\,=\, 20\Omega[/tex]

http://img397.imageshack.us/img397/9507/chapter2problem34part5su9.jpg [Broken]

Finally, adding the last two resistors in series, I get an [itex]R_{eq}[/itex] of [itex]40\Omega[/itex]. Does this seem correct?

http://img300.imageshack.us/img300/2751/chapter2problem34part6ma8.jpg [Broken]

Then, to find the Power, I get the current first.

[tex]i\,=\,\frac{v}{R}\,=\,\frac{12\,V}{40\Omega}\,=\,0.3\,A[/tex]

The I use the p = vi equation.

[tex]p\,=\,v\,i\,=\,\left(12\,V\right)\,\left(0.3\,A\right)\,=\,3.6\,W[/tex]

Right?
 
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  • #2
Or equivalently, P = V^2/R which gives the same answer.
 
  • #3


I would first commend you on your thorough and logical approach to solving this circuit. Your use of the correct equations for series and parallel resistors is commendable, and your final answer for equivalent resistance of 40 Ohms is correct.

To find the overall dissipated power, you correctly used the equation p = vi, where v is the voltage and i is the current. Your calculation of 3.6 W is also correct.

One thing to note is that since the voltage source is a DC source, the power dissipated in the circuit is constant. Therefore, you could have also used the formula p = i^2R or p = v^2/R to find the power, since the current and voltage are constant in this circuit. This may be a useful approach to consider in future circuit analysis problems.

Overall, your solution is correct and well thought out. Keep up the good work!
 

1. What is circuit analysis and why is it important?

Circuit analysis is the process of determining the behavior of electric circuits and their components. It is important because it allows us to understand how a circuit will behave and make predictions about its performance.

2. What is the equivalent resistance of a circuit?

The equivalent resistance of a circuit is the total resistance that a voltage source "sees" when connected to the circuit. It is a single resistance value that represents the combined effect of all the resistors in the circuit.

3. How do I find the equivalent resistance of a circuit?

To find the equivalent resistance of a circuit, you can use series and parallel resistance formulas or apply Kirchhoff's laws. You can also use Ohm's law to calculate the equivalent resistance by dividing the voltage by the current.

4. Why is finding the equivalent resistance important in circuit analysis?

Finding the equivalent resistance is important in circuit analysis because it allows us to simplify complex circuits and make calculations easier. It also helps us to determine the total amount of current flowing through the circuit and the voltage drops across each component.

5. Can the equivalent resistance of a circuit be greater than the sum of its individual resistances?

Yes, the equivalent resistance of a circuit can be greater than the sum of its individual resistances. This is because resistors in parallel have a combined effect that can decrease the total resistance, resulting in a lower equivalent resistance than the sum of the individual resistances.

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