Question

The primary of a step-down transformer has 240 turns and is connected to a 120 V rms line. The secondary is to supply 26 A at 6 V.
(a) Find the current in the primary.
_________A

(b) Find the number of turns in the secondary, assuming 100% efficiency.
_________

A 8 µF capacitor is charged to 33 V and is then connected across a 11-mH inductor.

(a) How much energy is stored in the system?
_________J

(b) What is the frequency of oscillation of the circuit?
________Hz

(c) What is the maximum current in the circuit?
_________A

Answer 1

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

$f=\frac{1}{2\pi \sqrt{LC}}$

$f=\frac{1}{2\times 3.14 \sqrt{0.011\times 8\times10^{-6}}}$

f = 536.78 Hz

Answer 2

Explanation:

Solution (1)

It is given that,

Number of turns in primary coil in step down transformer, $N_P=240$

Voltage in primary transformer, $V_p=120\ V$

Current in secondary coil, $I_s=26\ A$

Voltage in secondary coil, $V_s=6\ A$

(a) Let $I_P$ is the current in primary coil. It can be calculated using the formula as :

$\dfrac{V_P}{V_s}=\dfrac{I_s}{I_P}$

$\dfrac{120}{6}=\dfrac{26}{I_P}$

$I_p=1.3\ A$

(b) Let $N_s$ is the number of turns in secondary coil. It can be calculated using the formula as :

$\dfrac{V_P}{V_s}=\dfrac{N_P}{N_s}$

$\dfrac{120}{6}=\dfrac{240}{N_s}$

$N_s=12\ turns$

Solution (2)

Capacitance of the capacitor, $C=8\ \mu F=8\times 10^{-6}\ F$

Voltage, V = 33 V

Inductance, $L=11\ mH=11\times 10^{-3}\ H$

(a) Let E is the energy stored in the system. It is given by :

$E=\dfrac{1}{2}CV^2$

$E=\dfrac{1}{2}\times 8\times 10^{-6}\times (33)^2$

E = 0.00435 J

or

$E=4.35\times 10^{-3}\ J$

(b) Let f is the frequency of oscillation. It is given by :

$f=\dfrac{1}{2\pi \sqrt{LC} }$

$f=\dfrac{1}{2\pi \sqrt{11\times 10^{-3}\times 8\times 10^{-6}} }$

f = 536.51 Hz

(c) The maximum current in the circuit is,

$I=\sqrt{\dfrac{2E}{L}}$

$I=\sqrt{\dfrac{2\times 0.00435}{11\times 10^{-3}}}$

I = 0.88 A

Hence, this is the required solution.