Bryson Cowan

[Bryson Cowan]

yes the safety can get to L1 and N without going through another load

[Bryson Cowan]

P=I^2 X R so what i did was with I being amps and R being ohms i multiplied the amps by the total ohms in the circuit 6.67 amps X 36 ohms = 240 . 12

[Bryson Cowan]

3. the ignitor,booster,and main are in parallel and safety is in series

[Bryson Cowan]

i still came up with 240.12 watts

[Bryson Cowan]

1. wires carrying current always have inherent resistance or impedance to current flow voltage drop is defined as the amount of voltage loss that occurs through all or part of a circuit due to impedance. … this condition causes the load to work harder with less voltage pushing the current

2. it increases the current flow

3. the 1 load is in parallel and 2,3,and 4 are in series

[Bryson Cowan]

okay so i use the equation P=I^2 X R

[Bryson Cowan]

I MEANT P=I X R

[Bryson Cowan]

P=1XR?

[Bryson Cowan]

the equivelent resistance of two or more loads in parrellel will be something less than the smallest resistance

[Bryson Cowan]

okay that video helped out alot i understand it now so would the answer to this problem be [answer hidden] of equivelent resistance

• This reply was modified 3 years ago by Susan Brown.

[Bryson Cowan]

HW?

[Bryson Cowan]

yes i just rewhatched the video about it in unit 5 and im still having trouble understanding it i know what equation i need to use but im not sure how i need to work that equation if that makes any sense

[Bryson Cowan]

6.67 amps

[Bryson Cowan]

i multiplied the current by the resistance

[Bryson Cowan]

would it be 4 ohms of equivalent resistance?

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