Answer:
P = 68.125 kW
P startup = 43.05 kW
Explanation:
Determine the magnitude and sense (direction) of the current in the 500-latex: \omega ω resistor when i = 30 ma.
1 answer:
Complete Question:
Check the circuit in the file attached to this solution
Answer:
Total current = 0.056 A(From left to right)
Explanation:
Let the current in loop 1 be I₁ and the current in loop 2 be I₂
Applying KVL to loop 1
30 - (I₁ - I₂)500 + I₂R + 15 = 0
45 - 500I₁ - 500I₂ + RI₂ = 0
I₁ = 30mA = 0.03 A
45 - 500(0.03) - 500I₂ + RI₂ = 0
30 -500I₂ + RI₂ = 0...............(1)
Applying kvl to loop 2
-RI₂ - 15 + 10 - 400I₁ = 0
-RI₂ = 5 + 400*0.03
RI₂ = -17 ................(2)
Put equation (2) into (1)
30 -500I₂ -17 = 0
-500I₂ = 13
I₂ = -13/500
I₂ = -0.026 A
The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026
Total current = 0.056 A
The current will flow from left to right
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Explanation:
a) half-life of the radioactive substance:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
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