Determine the voltage drop from the top terminal to the bottom terminal, vab, in the right hand branch and, vcd, in the left han
1 answer:
Answer:
Hello your question is incomplete attached below is the missing part of the question
answer ;
voltage drop in the Vcd branch = 30 V
Voltage drop in the middle branch = 40v - 30v = 10 volts
voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts
Explanation:
Determine voltage drop from top terminal to bottom terminal ( Vab ) in the right hand branch and Vcd in left hand branch
40v and 50mA are in series hence; Ix = 50mA
also Vcd = 30V
CD is parallel to AB hence; Vcd = Vab = 30 V
Vab = ∝*Ix + 60 v
30v = ∝ ( 50mA ) + 60
therefore ∝ = -600
voltage drop in the Vcd branch = 30 V
Voltage drop in the middle branch = 40v - 30v = 10 volts
voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts
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Answer:
P_p = 27000 psf
Explanation:
given,
height of the retaining wall = h = 12 ft
internal angle of friction (∅)= 30°
unit weight = 125 pcf
Rankine passive earth pressure = ?
k_p is the coefficient of passive earth pressure
k_p = 3
Passive earth pressure
P_p = 27000 psf
Rankine passive earth pressure on the wall is equal to P_p = 27000 psf
Answer:
Explanation:
The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.
The answer to the above question is
The pressure at point 2 = 75.959 kPa
Explanation:
Bernoulli's equation with losses gives
hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)
Between points 1 and 2, z₁ = z₃ + 0.6 m therefore
hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)
hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)
But v = Q/A
or since A = π×D²/4 we have
A₁ = 1.77×10-2 m² , A₂ = 5.73×10-2 m², A₃ = 3.8×10-2 m²
Therefore from v = Q/A we have v₁ = 2.83 m/s v₂ = 0.87 m/s and v₃ = 1.315 m/s from there we find the friction coefficient from Moody Diagram as follows
ε = Which gives
the friction coefficients as f₁ = 0.02, f₂ = 0.017 and f₃ =0.0175
Substituting he above values into the equation we get
= 19.761 m
Combined head loss = 19.761 m
Hence 19.743 m = 0.6 m +(260 kPa-P₃)/(ρ×9.81) + (6.276)/(2×9.81)
or 260 kPa-18.82 m × 9.81 m/s²×ρ= P₃
Where ρ = density of water, we have
260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa