A tank circuit in a radio transmitter is a series RCL circuit connected to an antenna?
a asked:
A tank circuit in a radio transmitter is a series RCL circuit connected to an antenna?
A tank circuit in a radio transmitter is a series RCL circuit connected to an antenna. The antenna broadcasts radio signals at the resonant frequency of the tank circuit. Suppose that a certain tank circuit in a shortwave radio transmitter has a fixed capacitance of 2.4 x 10-11 F and a variable inductance. If the antenna is intended to broadcast radio signals ranging in frequency from 3.8 MHz to 9.1 MHz, find the (a) minimum and (b) maximum inductance of the tank circuit.
A tank circuit in a radio transmitter is a series RCL circuit connected to an antenna?
A tank circuit in a radio transmitter is a series RCL circuit connected to an antenna. The antenna broadcasts radio signals at the resonant frequency of the tank circuit. Suppose that a certain tank circuit in a shortwave radio transmitter has a fixed capacitance of 2.4 x 10-11 F and a variable inductance. If the antenna is intended to broadcast radio signals ranging in frequency from 3.8 MHz to 9.1 MHz, find the (a) minimum and (b) maximum inductance of the tank circuit.
When you apply AC to an inductor or capacitor in a circuit, they exhibit a property called reactance (X), measured in ohms. However, even though they are measured in the same units (ohms), each type of reactance is inversely related to each other so you have to call each inductive reactance (XL) and capacitive reactance (Xc).
Xc = 1/ (2 * PI * F * C)
where PI = 3.1416, F is frequency in Hertz and C is capacitance in farads. The result is capacitive reactance in ohms.
XL = 2 * PI * F * L
where PI and F are the same as above and L is inductance in Henries. The result is inductive reactance in ohms.
In your series RCL circuit, when inductive reactance equals capacitive reactance, they cancel out leaving you with just the resistance in the circuit. This also means the circuit is resonant. So all you have to do is set the two above reactance equations equal to each other and solve for L (inductance) at each of the two frequencies. You already know F and C.
1/ (2 * PI * F * C) = 2 * PI * F * L