Derive an expression for the resonant frequency of a parallel circuit, one branch consisting of a coil of inductance L and resistance R and the other branch of capacitance C.
Fig. 19
The series RL circuit with parallel C is given in Fig. 19. The designed circuit consists of resistance R, inductance L are in series with parallel configuration of capacitance C. Let us consider the current I flowing through the RLC circuit by the a.c. voltage source with constant value of V.
The current in the circuit
\(I=I_1+I_C\)
\(I=\frac{V}{R+j\omega L}+jV\omega C\)
\(I=V\left[\frac{1}{R+j\omega L}+j\omega C\right]\)
\(I=V\left[\frac{(R-j\omega L)}{R^2+{(\omega L)}^2}+j\omega C\right]\)
\(I=V\left[\frac{R}{R^2+{(\omega L)}^2}+j\left(-\frac{\omega L}{R^2+{(\omega L)}^2}+\omega C\right)\right]\)
Here at the resonance
\(\frac{\omega L}{R^2+{(\omega L)}^2}=\omega C\)
Or \(R^2+{(\omega L)}^2=\frac{L}{C}\)
Or \({(\omega L)}^2=\frac{L}{C}-R^2\)
\(\omega L=\sqrt{\frac{L}{C}-R^2}\)
The resonance frequency of the RLC circuit is \(f_0=\frac{1}{2\pi L}\sqrt{\frac{L}{C}-R^2}\)
A flux of 0.0006Wb is required in the air – gap of an iron ring of cross – section 5.0 cm2and mean length 2.7 m with an air – gap of 4.5 mm. Determine the ampere turns required. Six H values and corresponding B values are noted from the magnetization curve of iron and given below.
H(At/m) |
200 |
400 |
500 |
600 |
800 |
1000 |
B(Wb/m2) |
0.4 |
0.8 |
1.0 |
1.09 |
1.17 |
1.19 |
What is angle modulation?
Compare AM and NBFM.
Establish the equivalence between Thevenin’s and Norton’s theorems.
Draw and explain the square – law modulator.
Fig.23