Answer:
The circuit impedance is 3.84 phase 38.65º and the voltage across the capacitor is 0.13 phase -128.65º V.
Explanation:
Since the voltage given to us was Vs = 5*cos(5t) V and it is the form of V = Vmax*cos(omega*t) V we can extract the frequency omega, wich is w = 5 rad/s.
In the circuit we have a capacitor and a inductor. The capacitor impedance is negative and it is inversely proportional to the frequency, while the inductor impedance is positive and directly proportional to the frequency. So we have:
Z = R + jw*L - j/(wC)
Z = 3 + j*5*0.5 - j/(5*2)
Z = 3 + j*2.5 - j*0.1 = 3 + j*2.4 Ohm = 3.84 phase 38.65º Ohm
To find out the voltage across the capacitor we can use a voltage divider equation that is:
Vcapacitor = [Zcapacitor/(R + Zinductor + Zcapacitor)] * Vsource
Vcapacitor = [(-j0.1)/(3 + j2.4)]*Vsource
Vcapacitor = [(0.1 phase -90º)/(3.84 phase 38.65º)]*5 phase 0º
Vcapacitor = [0.026 phase -128.65º]* 5 phase 0º
Vcapacitor = 0.13 phase -128.65º V
Answer:
Explanation:
One the object is attached the toggle it's placed onto the bolt. The wings should be open in the direction of the bolt head. The toggle is than collapsed and the entire toggle bolt is inserted into the hole in the wall. Once the wings are fully through the opening they will spring open on the other side.
Hopt it's help thanks..
Where’s the illustration?
Answer:
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
Explanation:
The peak voltage after the 6 to 1 step down is . Then, the peak voltage of the rectified output is V_{d}[/tex] and according to the statement, the diodes can be modeled to be . Then, the peak voltage in the load is .
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
The average output voltage is calculated as:
The average current in the load is calculated as:
Answer:
I have attached the diagram for this question below. Consult it for better understanding.
Find the cross sectional area AB:
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
Forces is given by:
F = 2.75 × 10³ N
Horizontal Stress can be found by:
σ (x) = F/A
σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m
σ (x) = 143.23 × 10⁶ Pa
Horizontal Strain can be found by:
ε (x) = σ (x)/ E
ε (x) = 143.23 × 10⁶ / 200 × 10⁹
ε (x) = 716.15 × 10⁻⁶
Find Vertical Strain:
ε (y) = -v · ε (y)
ε (y) = -(0.3)(716.15 × 10⁻⁶)
ε (y) = -214.84 × 10⁻⁶
<h3>PART (a)</h3>
For L = 0.05m
Change (x) = L · ε (x)
Change (x) = 35.808 × 10⁻⁶m
<h3>
PART (b)</h3>
For W = 0.012m
Change (y) = W · ε (y)
Change (y) = -2.5781 × 10⁻⁶m
<h3>PART(c)</h3>
For t= 0.0016m
Change (z) = t · ε (z)
where
ε (z) = ε (y) ,so
Change (z) = t · ε (y)
Change (z) = -343.74 × 10⁻⁹m
<h3>
PART (d)</h3>
A = A(final) - A(initial)
A = -8.25 × 10⁻⁹m²
(Consult second picture given below for understanding how to calculate area)