Q5 (16 Marks) Electrical Circuits & Calculations
MET • Written Exam

(a) Explain the purpose of a rectifier and the effect of over current and of overvoltage on rectifiers. (6)

(b) Three equal resistors are connected to a three-phase system, If one resistor is removed find the reduction in load if they are connected in (i) Star, (ii) Delta. (10)

Appeared In: Aug 2026

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

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Part (a)

Purpose of a rectifier and effects of overcurrent and overvoltage:

  • A rectifier is a device (using diodes or thyristors) that converts alternating current (a.c.) into direct current (d.c.). It allows current to flow in one direction only, so the output is a unidirectional (pulsating) d.c. voltage. Rectifiers are used to supply d.c. for battery charging, excitation of generators, d.c. motors, and electronic equipment.
  • Effect of overcurrent: excessive current heats the semiconductor junction beyond its rated temperature, causing thermal breakdown and permanent damage to the diode/thyristor. It can also blow fuses, damage the transformer and cause the device to fail short-circuit. Overcurrent must be limited by fuses, circuit breakers or current-limiting circuits.
  • Effect of overvoltage: if the reverse voltage across a diode exceeds its peak inverse voltage (PIV) rating, the junction breaks down (avalanche) and conducts in the reverse direction, causing overheating and destruction. Overvoltage can also be caused by switching surges, lightning or inductive loads; surge suppressors (varistors, RC snubbers) are fitted to protect the rectifier. Both overcurrent and overvoltage must be limited to protect the semiconductor devices.
Part (b)

Three equal resistors on a three-phase system, one removed:

Let each resistor have resistance R and the line voltage be V.

(i) Star connection:

  • With three resistors in star, each phase voltage = V / root 3. Power per resistor = (V/root3)^2 / R = V^2 / (3R). Total power P3 = 3 x V^2/(3R) = V^2 / R.
  • With one resistor removed, two resistors remain in star. Power P2 = 2 x V^2/(3R) = 2V^2/(3R).
  • Reduction in load = P3 - P2 = V^2/R - 2V^2/(3R) = V^2/(3R).
  • Fractional reduction = (V^2/(3R)) / (V^2/R) = 1/3 = 33.3%.

(ii) Delta connection:

  • With three resistors in delta, each phase voltage = line voltage V. Power per resistor = V^2/R. Total P3 = 3V^2/R.
  • With one resistor removed, two resistors remain in delta. Power P2 = 2V^2/R.
  • Reduction = 3V^2/R - 2V^2/R = V^2/R.
  • Fractional reduction = (V^2/R)/(3V^2/R) = 1/3 = 33.3%.

So in both cases the load is reduced by one third (33.3%) when one of three equal resistors is removed.

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