Answer:
Z = 29.938Ω ∠22.04°
I = 2.494A
Explanation:
Impedance Z is defined as the total opposition to the flow of current in an AC circuit. In an R-L-C AC circuit, Impedance is expressed as shown:
Z² = R²+(Xl-Xc)²
Z = √R²+(Xl-Xc)²
R is the resistance = 4Ω
Xl is the inductive reactance = ωL
Xc is the capacitive reactance =
1/ωc
Given C = 12 μF, L = 6 mH and ω = 2000 rad/sec
Xl = 2000×6×10^-3
Xl = 12Ω
Xc = 1/2000×12×10^-6
Xc = 1/24000×10^-6
Xc = 1/0.024
Xc = 41.67Ω
Z = √4²+(12-41.67)²
Z = √16+880.31
Z = √896.31
Z = 29.938Ω (to 3dp)
θ = tan^-1(Xl-Xc)/R
θ = tan^-1(12-41.67)/12
θ = tan^-1(-29.67)/12
θ = tan^-1 -2.47
θ = -67.96°
θ = 90-67.96
θ = 22.04° (to 2dp)
To determine the current, we will use the relationship
V = IZ
I =V/Z
Given V = 12V
I = 29.93/12
I = 2.494A (3dp)
yes, a simple circuit will work without a switch. If we don't have a switch in a circuit then whichever device you are using electricity for wouldn't be able to switch off and will be working continuously. An open/close switch is usually all that is needed to supply or disconnect power from the source to the circuit.So, the circuit will work but it will work continuously.
Answer:
<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>
Explanation:
structural number = 4.5
stone base course material coefficient = 0.13
hma material layer coefficient = 0.40
drainage coefficient = 0.90
we will use layered analysis procedure to get thickness
D1 >= sN1/a1
when we cross multiply,
sN1 = a1D1 >=sN1
D2 >= -sN2-sN1/a2m2
sN2* + sN1* >= sN2
D3 >= sN3-(sN1*+sN2*)/a2m2
where a1,a2,a3 = layer coefficient
d1 d2 d3 = actual thickness
m2,m3 = coefficient of base
a1 = 0.4
a1 = 0.13
sN = 4.5
m2 = 0.9
D1 >= sN1/a1 = 4.5/0.4
= 11.25
thickness of surface = 12 inches
a1D1 = 0.4x12 = 4.8
we have value of sN2 = 5.5
(5.5 -4.8)/(0.13*0.9)
= 0.7/0.117
= 5.9829 inches
approximately 6 inches
so the pavement will have 12inch HMA surface and 6 inches granular base layer.
