The amplitude current of the RLC series AC circuit is determined as 11.1 mA.
How to calculate the amplitude current?
The capacitance of the capacitor is calculated as follows;
Xc = 1/ωC
where;
- ω is the angular speed
- C is capacitance
Xc = 1/(405 x 4.95 x 10⁻⁶)
Xc = 498.82 Ω
The inductance of the inductor is calculated as;
Xl = ωL
Xl = (405 x 0.409)
Xl = 165.65 Ω
The impedance of the circuit is calculated as follows;
Z = √ (R² + (Xc - Xl)² )
Z = √ (192² + (498.82 - 165.65)² )
Z = 384.53 Ω
The root mean square current is calculated as;
Irms = V/Z
Irms = 3.02 / 384.53
Irms = 0.00785 A
The amplitude current is calculated as follows;
Irms = 0.7071 × I₀
I₀ = Irms / 0.7071
I₀ = 0.00785 A / 0.7071
I₀ = 0.0111 A
I₀ = 11.1 mA
The complete question is below:
You have a 192 Ω resistor, a 0.409 H inductor, a 4.95 μF capacitor, and a variable-frequency ac source with an amplitude of 3.02 V. You connect all four elements together to form a series circuit.
what will be the current amplitude at an angular frequency of 405 rad/s ? express your answer with the appropriate units.