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3 years ago

Find the current in 1 the electrical circuit below

Engineering

1 answer:

Lesechka [4]3 years ago

3 0

Answer:

c. 25 μA

Explanation:

The dependent current source means that 31 times i1 will flow through the 6kΩ resistor, effectively increasing its voltage drop to 31 times that which would result from i1 alone. In effect, the 6kΩ resistor behaves in the left-side circuit loop as though it were 31×6kΩ = 186kΩ (with no dependent current source).

Then the current i1 is equivalent to that created by a 5+1 = 6V source through a 54kΩ +186kΩ = 240kΩ circuit impedance.

(6V)/(240kΩ) = 25 μA

_____

<em>Additional comment</em>

The voltage across the 6kΩ resistor is (186/240)·6V = 4.65V, and the 25 μA current generates a voltage of 30·(25 μA)(1.8kΩ) = 1.35V across the 1.8kΩ resistor. This means the voltage source at the right side of the diagram needs to be at least 4.65 +1.35 = 6.0V in order to support the calculated voltage drops.

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3 years ago

How high a building could fire hoses effectively spray from the ground? Fire hose pressures are around 1 MPa. (It is also said t

Mrac [35]

Answer:

$z_{2} = 91.640\,m$

Explanation:

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The velocity of water delivered by the fire hose is:

$v_{1} = \frac{(300\,\frac{gal}{min} )\cdot(\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot (0.3\,m)^{2}}$

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$z_{2}= \frac{P_{1}-P_{2}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}$

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