(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.
Verified Correct Answer: .
Physical Principles & Numerical Solution
(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)
Source: Myschool Past Questions Archive (Record #54094)