GATE 2018 EE Q.49: solution and answer

GATE 2018 · EEQ.49 · 2 marks · NAT

A phase controlled single phase rectifier, supplied by an AC source, feeds power to an R-L-E load as shown in the figure. The rectifier output voltage has an average value given by $V_{o} = \frac{V _{m}}{2 π} (3 + cos α)$ , where $V_{m} = 80 π$ volts and $α$ is the firing angle. If the power delivered to the lossless battery is 1600 W, $α$ in degree is ________ (up to 2 decimal places).

GATE 2018 EE question 49: Source V_m sin (ω t) on the left feeds a rectifier block (a box containing a single thyristor symbol with a gate terminal). — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

ANSWER

The firing angle is $α = 90.0 0^{\circ}$ .

Official answer: 90.0Source: Official answer key, GATE 2018 EE (IIT Guwahati)

The battery is lossless, so all the power it receives is $P = E I_{o}$ with $E = 80 V$ . Hence $I_{o} = \frac{1600}{80} = 20 A$ (the average load current).
In steady state the average voltage across the 10 mH inductor is zero, so only the resistor and the battery drop average voltage: $V_{o} = E + I_{o} R = 80 + 20 \times 2 = 120 V$ .
The given average output voltage is $V_{o} = \frac{V _{m}}{2 π} (3 + cos α)$ with $V_{m} = 80 π V$ , so $V_{o} = \frac{80 π}{2 π} (3 + cos α) = 40 (3 + cos α) V$ .
Equate the two expressions for $V_{o}$ : $40 (3 + cos α) = 120 \Rightarrow 3 + cos α = 3 \Rightarrow cos α = 0$ .
The firing angle is therefore $α = cos^{- 1} (0) = 9 0^{\circ}$ , that is $90.0 0^{\circ}$ .

KEY IDEA

The power into a lossless battery is E times the average current. Because the inductor has zero average voltage in steady state, the average output voltage is the battery voltage plus the resistor drop, and this is equated to the given expression for the average output voltage.

WATCH OUT FOR

Including the inductor in the average-voltage balance, or taking the power as $I_{r m s}^{2} R$ or $V_{o} I_{o}$ . The 1600 W is the battery's power only, so it equals $E I_{o}$ . The resistor loss is extra and is not part of it.

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 12 Rectifiers questions, with notes

GATE 2018 EE Q.49: solution and answer

Answer and solution

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