Q9 (10 Marks) Electrical Circuits & Calculations
MET • Written Exam

(a) What is the operational impedance of an R.C. Circuit? Describe its usefulness. (6)

(b) A ring-main, 900m long is supplied at a point A at a p.d. of 220V. At a point B, 240m from A, a load of 45A is drawn from the main, and at a point C, 580m from A, measured in the some direction, a load of 78A is taken from the main, if the resistance of the main (lead and return) is 0.25 ohm per kilometre, calculate the current which will flow in each direction round the main from the supply point A and the potential difference across the main, at the load where it is lowest?

Appeared In: Nov 2023

Verified Model Answer (Text Solution)

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Part (a)

Operational impedance of an R.C. circuit and its usefulness:

  • The operational impedance of an R.C. circuit is the total opposition to current, combining the resistance R and the capacitive reactance XC = 1/(2 pi f C). For a series R-C circuit, Z = R - j XC, with magnitude |Z| = sqrt(R^2 + XC^2) and phase angle phi = atan(XC/R) (current leading voltage).
  • Usefulness: it determines the current and phase angle in the circuit for a given voltage and frequency, and hence the power (P = V I cos phi). It is used to design filters, timing circuits, coupling circuits, and to analyse the behaviour of circuits containing capacitors. The impedance shows how the circuit responds to frequency (a capacitor blocks d.c. and passes high frequencies).
Part (b)

Ring main, 900 m long, supplied at A at 220 V. Load 45 A at B (240 m from A), load 78 A at C (580 m from A). Resistance 0.25 ohm/km (lead and return).

  • Resistance per metre = 0.00025 ohm/m.
  • Segment resistances: A-B = 0.06 ohm; B-C = 0.085 ohm; C-A (closing) = 0.08 ohm.
  • Let x = current from A towards B, y = current from A the other way to C. x + y = 123 A.
  • Current in A-B = x; in B-C = x - 45; in short path A-C = y.
  • Loop equation: 0.06 x + 0.085(x - 45) = 0.08 y.

0.145 x - 3.825 = 0.08(123 - x) -> 0.225 x = 13.665 -> x = 60.73 A.

  • y = 123 - 60.73 = 62.27 A.
  • Voltage at B: 220 - 0.06 x 60.73 = 220 - 3.64 = 216.36 V.
  • Voltage at C: 220 - 0.08 x 62.27 = 220 - 4.98 = 215.02 V.
  • The lowest voltage is at C: 215.0 V.

So currents from A are 60.7 A (towards B) and 62.3 A (towards C), and the lowest voltage is about 215 V at load C.

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