Answer:
The reflected resistance in the primary winding is 6250 Ω
Explanation:
Given;
number of turns in the primary winding, = 50 turns
number of turns in the secondary winding, = 10 turns
the secondary load resistance, = 250 Ω
Determine the turns ratio;
Now, determine the reflected resistance in the primary winding;
Therefore, the reflected resistance in the primary winding is 6250 Ω
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