What is the Power Loss in a 20 Ohm Resistor in a Circuit?

• poopoo16
In summary, the conversation discussed how to calculate the power loss in a 20 ohm resistor in a given circuit. It was determined that the 6 ohm resistor needed to be taken into account and could be treated as part of a series with the 4 ohm resistor. The current flowing through the circuit was found to be 1 amp, and the next step was to calculate the voltage across the 20 ohm and 5 ohm resistors. This could be done using the total current and equivalent resistance, or by calculating the voltage drop across the 6 ohm resistor and subtracting it from the source voltage. The conversation then discussed options for calculating the power using either voltage or current equations.
poopoo16
ok consider the circuit. what is the power loss in the 20 ohm resistor in J/s?

ok so what i did was that... and i don't know where to go after... please help

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poopoo16 said:
ok consider the circuit. what is the power loss in the 20 ohm resistor in J/s?

ok so what i did was that... and i don't know where to go after... please help

That's a good start. Now what about the 6 ohm resistor?

Reduce the circuit to an equivalent resistor, find the flowing current, and from there use your power equations to find the power.

You are on the right track.

OlderDan said:
That's a good start. Now what about the 6 ohm resistor?

that s my question... don't know how to treat the 6 ohms... do i add it to the 4?

and then use P=IV where I=V/R?...if i do that, its give me 1...and that is not the right answer...

poopoo16 said:
that s my question... don't know how to treat the 6 ohms... do i add it to the 4?

Yes, because it's in series with the other two. Be careful, however, in the power calculation. Is the current flowing through the 20 ohm resistor the same as through the whole circuit?

well, after u combine the 2 parallel ones, its no longer going thru the 20 per se right, its going thru a 6 and a 4?...soo... it then a series right?

so...

R=10
V=10
I=V/R
hence I=10/10?
meanning
P=IV
P=(1)10...?

poopoo16 said:
that s my question... don't know how to treat the 6 ohms... do i add it to the 4?

and then use P=IV where I=V/R?...if i do that, its give me 1...and that is not the right answer...

What you have found IS the right answer for the current flowing through the power source and through the 6 ohm resistor. The next step is to figure out the voltage across the 20 ohm and 5 ohm resistors (both the same). You can do that by using the total current and the equivalent resistance of that parallel combination that you found in the first step, or you can calculate the voltage drop across the 6 ohm resistor when that 1 amp of current flows through it and subtract that from the 10 volt source voltage. Once you have that voltage, you can use a power equation in terms of V and R, or you can use it to calculate the current through each of those resistors and use the power equation in terms of I and V. You should note that the 1 amp of current gets divided into two parts that flow through the two paths (5 ohm and 20 ohm) and that the sum of those two parts must add up to the current flowing through the 6 ohm and the power source, the 1 amp you have already found.

how do i calcuate voltyage drop arcross the 6ohm? do i use P=I^2R?

You used V = IR solved for I to find the current of 1amp. Now you know the current through the 6 ohm resistor. Use V = IR to find the voltage across that one resistor.

1. What is power loss in a circuit?

Power loss in a circuit refers to the decrease in the amount of power that is available to the load due to various factors such as resistance, impedance, and other inefficiencies in the circuit.

2. How does power loss occur in a circuit?

Power loss can occur in a circuit due to several reasons, including resistance in the wires, connections, and components, as well as heat dissipation and electromagnetic interference.

3. How can power loss be calculated in a circuit?

Power loss can be calculated by multiplying the resistance in the circuit by the square of the current flowing through it. The formula for power loss is P = I^2 * R, where P is the power loss, I is the current, and R is the resistance.

4. What are some ways to reduce power loss in a circuit?

To reduce power loss in a circuit, you can use larger wire sizes with lower resistance, minimize connections to reduce resistance, and use components with lower resistance values. Additionally, proper circuit design and regular maintenance can also help reduce power loss.

5. How does power loss affect the performance of a circuit?

Power loss can significantly impact the performance of a circuit by reducing the amount of power available to the load. This can result in voltage drops, reduced current, and ultimately affect the functionality of the circuit.

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