Respuesta :
Explanation:
Solution (1)
It is given that,
Number of turns in primary coil in step down transformer, [tex]N_P=240[/tex]
Voltage in primary transformer, [tex]V_p=120\ V[/tex]
Current in secondary coil, [tex]I_s=26\ A[/tex]
Voltage in secondary coil, [tex]V_s=6\ A[/tex]
(a) Let [tex]I_P[/tex] is the current in primary coil. It can be calculated using the formula as :
[tex]\dfrac{V_P}{V_s}=\dfrac{I_s}{I_P}[/tex]
[tex]\dfrac{120}{6}=\dfrac{26}{I_P}[/tex]
[tex]I_p=1.3\ A[/tex]
(b) Let [tex]N_s[/tex] is the number of turns in secondary coil. It can be calculated using the formula as :
[tex]\dfrac{V_P}{V_s}=\dfrac{N_P}{N_s}[/tex]
[tex]\dfrac{120}{6}=\dfrac{240}{N_s}[/tex]
[tex]N_s=12\ turns[/tex]
Solution (2)
Capacitance of the capacitor, [tex]C=8\ \mu F=8\times 10^{-6}\ F[/tex]
Voltage, V = 33 V
Inductance, [tex]L=11\ mH=11\times 10^{-3}\ H[/tex]
(a) Let E is the energy stored in the system. It is given by :
[tex]E=\dfrac{1}{2}CV^2[/tex]
[tex]E=\dfrac{1}{2}\times 8\times 10^{-6}\times (33)^2[/tex]
E = 0.00435 J
or
[tex]E=4.35\times 10^{-3}\ J[/tex]
(b) Let f is the frequency of oscillation. It is given by :
[tex]f=\dfrac{1}{2\pi \sqrt{LC} }[/tex]
[tex]f=\dfrac{1}{2\pi \sqrt{11\times 10^{-3}\times 8\times 10^{-6}} }[/tex]
f = 536.51 Hz
(c) The maximum current in the circuit is,
[tex]I=\sqrt{\dfrac{2E}{L}}[/tex]
[tex]I=\sqrt{\dfrac{2\times 0.00435}{11\times 10^{-3}}}[/tex]
I = 0.88 A
Hence, this is the required solution.
Answer:
Explanation:
Np = 240
Vp = 120 V
Is = 26 A
Vs = 6 V
(a) According to the principle of transformer
Ip/ Is = Vs / Vp
Ip / 26 = 6 / 120
Ip = 1.3 A
(b) Ns / Np = Vs / Vp
Ns / 240 = 6 / 120
Ns = 12
....................................
C = 8 x 10^-6 F
V = 33 V
L = 11 m H = 0.011 H
(a) Energy stored, U = 0.5 x CV² = 0.5 x 8 x 10^-6 x 33 x 33
U = 4.356 x 10^-3 J
(b) Let f be the frequency of oscillations
[tex]f=\frac{1}{2\pi \sqrt{LC}}[/tex]
[tex]f=\frac{1}{2\times 3.14 \sqrt{0.011\times 8\times10^{-6}}}[/tex]
f = 536.78 Hz
