Answer:
Its inductance L = 166 mH
Explanation:
Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR
R = V/I = 5.62/0.698 = 8.052 Ω
Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from
V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'
Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω
WE now find the reactance X of the coil from
Z² = X² + R²
X = √(Z² - R²)
= √(97.5² - 8.05²)
= √(9506.25 - 64.8025)
= √9441.4475
= 97.17 Ω
Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.
L = X/2πf
= 97.17/2π(93.1 Hz)
= 97.17 Ω/584.965 rad/s
= 0.166 H
= 166 mH
Its inductance L = 166 mH
Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given that,
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;
Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Answer:
970 kN
Explanation:
The length of the block = 70 mm
The cross section of the block = 50 mm by 10 mm
The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN
The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN
By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force
The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa
The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa
The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa
The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN
Its false I got It right .
Answer:
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