(a) Define: (i) reactance; (ii) impedance.
(b)(i) Explain resonant frequency of an RLC circuit.
(ii) Explain the statement — the power supply voltage of a source is 230V
(c) A source of e.m.f 240V and frequency 50 Hz is connected to a series arrangement of a resistor, an inductor and a capacitor. When the current in the capacitor is 10A, the potential difference across the resistor is 140 V and that across the inductor is 50 V. Calculate the:
(i) potential difference across the capacitor (ii) capacitance of the capacitor; (iii) inductance of the indicator.
(d) Draw and label one vector diagram for the potential differences across the inductor, the capacitor and the resistor in (c) above.
Explanation
(a)(i) Reactance s the opposition in ohms to the alternating current by an inductor, a capacitor or both.
(ii) Impedance is the total opposition is ohms to the alternating current by a combination of resistor as well as a capacitor or an inductor.
(b)(i) Resonant frequency fo is to frequency at which the current in the circuit has a maximum value. fo = (frac{1}{2 pi LC})
(ii) This means that the root mean square value of the voltage is 230v. The actual voltage lies between Vo and -Vo where the voltage amplitude or peak voltage Vo.
(c)(i) (V^2 = V^2_R + (V_L – V_C)^2; 240^2 = 140^2 + (50 – v_c)^2);
50 – (V_c = 194.9)
(V_c = 244.9v)
(ii) X(_c = frac{1}{2 pi fc} = frac{V}{1} = frac{1}{2 pi times 50 times c} = frac{244.9}{10})
C = (frac{10}{2 pi times 50 times 244.9} = 129.97 mu F)
(iii) (X_L = 2 pi fL = frac{v}{l}; 2 pi times 50 times L = frac{50}{10})
L = (frac{50}{100x})
= 0.01159Hz
(d)