The figure below shows a half-bridge voltage source inverter supplying an RL-load with R = 40 Ω and \(L = \left( {\frac{{0.3}}{\pi }} \right)H\). The desired fundamental frequency of the load voltage is 50 Hz. The switch control signals of the converter are generated using sinusoidal pulse width modulation with modulation index M = 0.6. At 50 Hz, the RL-load draws an active power of 1.44 kW. The value of DC source voltage V_{DC} in volts is

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GATE EE 2017 Official Paper: Shift 2

Option 3 : \(500\sqrt 2\)

CT 1: Ratio and Proportion

3742

10 Questions
16 Marks
30 Mins

__Concept__:

From sinusoidal PWM control technique:

The peak value of fundamental voltage is given as,

Vm1 = MVdc ---(1)

__ Explanation__:

Given that,

RL load: R = 40 Ω, \(L = \frac{0.3}{π} \) H

f = 50 Hz

P = 1.44 kW

Modulation index (M) = 0.6

At 50 Hz,

XL = 2πfL = 100 × \(\frac{0.3}{\pi} \) = 30 Ω

Z = R + jXL = 40 + j30

|Z| = 50 Ω

∵ P = 1.44 × 103

⇒ \(\rm I_{rms}^2 R = 1440\)

\(⇒ \rm I_{rms}^2 R = \frac{1440}{40}\)

⇒ Irms = 6 A

we know that,

\(\rm Z = \frac{V_{rms}}{I_{rms}}\)

⇒ Vrms = Irms × Z

⇒ Vrms = 60 × 50

⇒ Vrms = 300 V

so, peak value Vm = 300√2

From equation (1)

300√2 = 0.6 VDC

⇒ VDC = \(\rm \frac{300√ 2}{0.6}\)

∴ VDC = 500√2 V