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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
OlderDan said:That's a good start. Now what about the 6 ohm resistor?
poopoo16 said:that s my question... don't know how to treat the 6 ohms... do i add it to the 4?
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...
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.
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.
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.
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.
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.