How is the total admittance Y of a parallel R-L-C circuit formed?

How is the total admittance Y of a parallel R-L-C circuit formed?

• A. Y = 1/R
• B. Y = 1/(R + jωL) + jωC
• C. Y = 1/R + jωL + 1/(ωC)
• D. Y = 1/R + 1/(jωL) + jωC

Answer: (D)

In a parallel circuit, the total admittance is simply the sum of the admittances of each branch. For each element: - Resistor: Y = 1/R (purely real). - Inductor: Z = jωL, so Y = 1/Z = 1/(jωL) = -j/(ωL) (purely imaginary negative). - Capacitor: Z = 1/(jωC), so Y = 1/Z = jωC (purely imaginary positive). Adding these gives Y = 1/R + 1/(jωL) + jωC. This shows how the capacitor adds a positive imaginary component, the inductor a negative imaginary component, and the resistor only a real component. The other forms mix up series vs. parallel relationships or misuse the impedance-to-admittance conversions.

In a parallel circuit, the total admittance is simply the sum of the admittances of each branch. For each element:

• Resistor: Y = 1/R (purely real).

• Inductor: Z = jωL, so Y = 1/Z = 1/(jωL) = -j/(ωL) (purely imaginary negative).

• Capacitor: Z = 1/(jωC), so Y = 1/Z = jωC (purely imaginary positive).

Adding these gives Y = 1/R + 1/(jωL) + jωC. This shows how the capacitor adds a positive imaginary component, the inductor a negative imaginary component, and the resistor only a real component. The other forms mix up series vs. parallel relationships or misuse the impedance-to-admittance conversions.