Shortcut Methods
JEE Mains:
-
Series LCR Circuit:
-
Power factor (P.F) = Cos θ = R/Z
-
Impedance Z = √(R^2 + (Xl - Xc)^2) where R = 10 Ω, XL (inductive reactance) = 2πfL = 2 × π × 50 × 0.2 = 62.83 Ω XC (capacitive reactance) = 1/(2πfC) = 1/(2 × π × 50 × 200 × 10^-6) = 159.15 Ω
-
Z = √(10^2 + (62.83 - 159.15)^2) = 162.47 Ω Therefore, P.F. = Cos θ = 10/162.47 = 0.0616
-
Parallel LCR Circuit:
-
Admittanc (Y) = √(G^2 + (B - C)^2) where G = 1/R = 1/10 = 0.1 S B (susceptance due to inductance) = 2πfL = 2 × π × 50 × 0.2 = 62.83 S C (susceptance due to capacitance) = 2πfC = 2 × π × 50 × 200 × 10^-6 = 159.15 S
-
Y = √(0.1^2 + (62.83 - 159.15)^2) = 162.47 S
-
P.F. = Cos θ = G/Y = 0.1/162.47 = 0.000616
-
Series LCR Circuit with Voltage, Current, and Frequency:
- Power factor (P.F) = Cos θ = R/(V/I) = R x I/V = 10 × 1/100 = 0.1
CBSE Board Exams:
-
Series LCR Circuit:
- Power (P) = I^2 x R P = (1)^2 x 10 = 10 Watts
-
Parallel LCR Circuit:
- Impedance Z = √(R^2 + (Xl - Xc)^2) = 162.47 Ω
- Current I = V/Z = 100/162.47 = 0.616 A
- Power (P) = I^2 x R = (0.616)^2 x 10 = 3.82 Watts
-
Series LCR Circuit with Voltage, Current, and Frequency:
-
Power (P) = V x I x Cos θ P = 100 x 1 x 0.1 = 10 Watts
