To solve the problem of finding the impedance, resistance, capacitive reactance, and inductive reactance in an LCR series circuit, we follow these steps:
Thus, the values are:
(A) Impedance = 150Ω,
(B) Resistance = 120Ω,
(C) Capacitive Reactance = 100Ω,
(D) Inductive Reactance = 10Ω.
Arranging the values in increasing order: (D), (C), (B), (A).
Given LCR circuit parameters:
E = 30sin(100t) V, R = 120 Ω, L = 100 mH, C = 100 μF
1. Angular frequency (ω): \[ ω = 100\ \text{rad/s} \] (from the source equation)
2. Inductive reactance (XL): \[ X_L = ωL = 100 × 0.1 = 10\ \Omega \]
3. Capacitive reactance (XC): \[ X_C = \frac{1}{ωC} = \frac{1}{100 × 100×10^{-6}} = 100\ \Omega \]
4. Impedance (Z): \[ Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{120^2 + (100 - 10)^2} \] \[ Z = \sqrt{14400 + 8100} = \sqrt{22500} = 150\ \Omega \]
Values obtained:
A. Impedance (Z) = 150 Ω
B. Resistance (R) = 120 Ω
C. Capacitive reactance (XC) = 100 Ω
D. Inductive reactance (XL) = 10 Ω
Increasing order: D (10 Ω) < C (100 Ω) < B (120 Ω) < A (150 Ω)
Final answer: The values in increasing order are D < C < B < A.
The Wheatstone bridge is an arrangement of four resistances, say \(R_1, R_2, R_3\), and \(R_4\). The null point condition is given by: