The function of those little black transformers that clog your power strips is to step down line voltage to run small electronic gadgets. Suppose a fax machine uses an ideal transformer with a primary voltage of 120 V and primary windings. If the secondary voltage is 12 V find
A.) The number of turns on the secondary
B.) If the fax machine at idle draws enough current to dissipate 6 W, find the current that must flow in the primary.
Find the resonant frequency of a circuit that has a 400 mH inductor and a 30 nF capacitor connected in series.
A. Np/Ns = Vp/Vs = 120/12 = 10/1.
Turns Ratio = 10:1
Ts = Tp/10
To find the number of turns on the secondary(Ts), we need to know the
number of turns on the primary(Tp).
B. 120 * I = 6 W. Solve for I.
Fr = 1/(2pi*Sqrt(LC)
Fr = 1(6.28*Sqrt( 0.40*30*10^-9) =
8.33 Million Hertz = 8.33 mHz.
A.) To find the number of turns on the secondary of an ideal transformer, we need to use the turns ratio formula, which states that the turns ratio is equal to the voltage ratio. In this case, the primary voltage is 120 V and the secondary voltage is 12 V.
Turns ratio = Secondary voltage / Primary voltage
Turns ratio = 12 V / 120 V
Turns ratio = 1/10
Since the turns ratio is the ratio of the number of turns on the secondary to the number of turns on the primary, we can say:
1/10 = Number of turns on the secondary / Number of turns on the primary
To find the number of turns on the secondary, we need to multiply the number of turns on the primary by the turns ratio:
Number of turns on the secondary = Number of turns on the primary * Turns ratio
Number of turns on the secondary = Number of turns on the primary * 1/10
Therefore, the number of turns on the secondary is one-tenth of the number of turns on the primary.
B.) To find the current that must flow in the primary of the fax machine, we need to use the power formula, which states that power is equal to the current squared multiplied by the resistance. In this case, we have a power dissipation of 6 W.
Power = Current^2 * Resistance
Since we know the power (6 W) and we have an ideal transformer with no resistance, we can rearrange the formula to solve for the current:
Current^2 = Power / Resistance
Current^2 = 6 W / 0 Ω
Since the resistance is zero (ideal transformer), the current will be infinite. However, in practical scenarios, transformers are not ideal and have some impedance, which limits the current.
To summarize, the number of turns on the secondary is one-tenth of the number of turns on the primary, and the current that must flow in the primary is dependent on the resistance of the transformer, which is typically not zero.
Moving on to the second question:
To find the resonant frequency of a circuit that has a 400 mH inductor and a 30 nF capacitor connected in series, we can use the formula for calculating resonant frequency in an LC circuit:
Resonant frequency = 1 / (2π√(Inductance * Capacitance))
Resonant frequency = 1 / (2π√(400 mH * 30 nF))
First, let's convert the units consistently:
Resonant frequency = 1 / (2π√(0.4 H * 0.00003 F))
Resonant frequency = 1 / (2π√(0.000012 H * F))
Resonant frequency = 1 / (2π√(1.2 x 10^-5 H * F))
Resonant frequency = 1 / (2π√(1.2 x 10^-5))
Resonant frequency = 1 / (2π * √(1.2 x 10^-5))
Finally, we can calculate the resonant frequency using a calculator or a computer program:
Resonant frequency ≈ 44955.01 Hz (rounded to five decimal places)
So, the resonant frequency of the circuit with a 400 mH inductor and a 30 nF capacitor connected in series is approximately 44.955 kHz.