Power Dissipation in Circuit Calculation

In summary, the power dissipated by the 5.0Ω resistor in the given circuit is 3.46W. This is calculated by finding the voltage drop across the resistor, which is 4.15V, and using that with the current of 0.833333A to find the power (P=IV). The confusion about voltage in parallel circuits is clarified by understanding that the voltage drop across loads connected in parallel is constant, while in series it is proportional to the resistance (V=IR).
  • #1
brentwoodbc
62
0

Homework Statement



What is the power dissipated by the 5.0Ω resistor in the following circuit?
4000123.gif

Homework Equations

I thought maybe to simplify the resistance like.

1/r=1/30 +1/30
r=15
+10+5
R=30ohms
then find the current
v=ir
25=Ix30
I=.8333
then find power with the 5 ohms?

P=vi
P=25x5
P=75w?
answer given is 6.2w?

thanks.
 
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  • #2
Your current calculations look ok.

You have used the wrong value for voltage.

Remember that only a proportion of the 25V is used in the 5 ohm resistor.

I'll leave it up to you to figure out what the value should be.

Oh, and you've substituded wrongly also...
 
Last edited:
  • #3
You can also find a formula for power in terms of current and resistance, instead of current and voltage as you have it.
 
  • #4
Im trying to think of what the voltage is at that point,
do I use,
EMF=Vab+IR ?
24= vab+0.83*5 or something?

I thought voltage was constant everywhere in a circuit?
 
Last edited:
  • #5
The voltage drop across loads connected in parallel is constant(equal to supplied voltage) whereas the voltage drop across loads connected in series is proportional to their resistance (V=IR)

So, calculate the voltage drop across the 5Ohm resistor and use that in you're calculation for Power.

Edit: I got 3.47W as the power dissipation though... I might have missed something.
 
  • #6
Jiachao said:
The voltage drop across loads connected in parallel is constant(equal to supplied voltage) whereas the voltage drop across loads connected in series is proportional to their resistance (V=IR)

So, calculate the voltage drop across the 5Ohm resistor and use that in you're calculation for Power.

Edit: I got 3.47W as the power dissipation though... I might have missed something.

your right its 3.47, the question was split over 2 pages so I saw the wrong answer.

I'm still kind of confused about the voltage when in parallel though.
so if you have a 20v power source that splits 4 ways each is 20v then as far as resistance you treat each on seperatly as if it was a series?
 
  • #7
Thanks.

so v=IR
25=Ix30
I=.833333

V=IR
V=.83x5
V=4.15

P=IV
P=4.15x.8333333
P=3.46Is that right?
 
  • #8
Yes, that's how I did it.
 

1. What is power dissipation in circuit calculation?

Power dissipation in circuit calculation refers to the process of determining the amount of power that is lost or dissipated within a circuit. This is an important factor to consider as it affects the overall efficiency and performance of a circuit.

2. How is power dissipation calculated in a circuit?

Power dissipation is calculated by multiplying the voltage across a component by the current flowing through it. This can be done for each individual component in a circuit and then added together to determine the total power dissipation.

3. What factors affect power dissipation in a circuit?

There are several factors that can affect power dissipation in a circuit, including the resistance of the components, the current flowing through the circuit, and the temperature of the components. Additionally, the type of circuit and its design can also impact power dissipation.

4. Why is power dissipation important to consider in circuit design?

Power dissipation is an important consideration in circuit design because it can affect the overall efficiency and performance of a circuit. High levels of power dissipation can lead to overheating and potentially damage the components in a circuit. It is also important for energy conservation and cost-effectiveness.

5. How can power dissipation be reduced in a circuit?

There are several ways to reduce power dissipation in a circuit, such as using components with lower resistance, optimizing the circuit design to minimize current flow, and implementing cooling methods to reduce the temperature of the components. Additionally, using energy-efficient components and properly sizing the circuit can also help reduce power dissipation.

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