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A single phase half bridge inverter has RLC load. The dc input voltage (Vs/2) = 115 V and the output frequency is 50 Hz. The expression for the load voltage up to the fifth harmonic will be given by
b Explanation: In a single phase HALF bridge inverter only odd harmonics are present. i.e. 1,3,5 etc. Vo = (2Vs/π) sin ωt + (2Vs/3π) sin 3ωt + (2Vs/5π) sin 5ωt . . . (2Vs/π) = 146 V ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t
b
Explanation: In a single phase HALF bridge inverter only odd harmonics are present. i.e.
1,3,5 etc.
Vo = (2Vs/π) sin ωt + (2Vs/3π) sin 3ωt + (2Vs/5π) sin 5ωt . . .
(2Vs/π) = 146 V
ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t
See lessA single phase full bridge inverter has RLC load with R = 4 Ω, L = 35 mH and C = 155 μF. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the angle by which the third harmonic current will lead/lag the third harmonic output voltage.
d Explanation: X L = 2 x 3.14 x 50 x 0.035 = 10.99 Ω X C = 1/(2 x 3.14 x 50 x 155 x 10 -6 ) = 20.54 Ω For the third harmonic component X L (3rd harmonic) = 10.99 x 3 = 33 Ω (approx.) X C (3rd harmonic) = 20.54/3 = 6.846 Ω R = 4 Ω P = tan -1 (X L – X C )/R = 81.3°
d
Explanation:
X
L = 2 x 3.14 x 50 x 0.035 = 10.99 Ω
X
C = 1/(2 x 3.14 x 50 x 155 x 10
-6
) = 20.54 Ω
For the third harmonic component
X
L
(3rd harmonic) = 10.99 x 3 = 33 Ω (approx.)
X
C
(3rd harmonic) = 20.54/3 = 6.846 Ω
R = 4 Ω
P = tan
-1
(X
L – X
C
)/R = 81.3°
See lessA single phase full bridge inverter has RLC load with R = 4 Ω, Xl = 11 Ω and Xc = 20.54 Ω. The dc input voltage is 230 V. Find the value of fundamental load power.
b Explanation: Average value of fundamental output voltage = 4Vs/π = 292.85 V Z = [R 2 + (Xl – Xc) 2 ] 1/2 I = V/Z = 28.31 I(rms)(fundamental) = V/Z√2 = (292.85)/(1.414 x 10.345) = 20.02 A Fundamental Load power = (20.02) 2 x R = (20.02) 2 x 4 = 1603.2 Watts.
b
Explanation: Average value of fundamental output voltage = 4Vs/π = 292.85 V
Z = [R
2 + (Xl – Xc)
2
]
1/2
I = V/Z = 28.31
I(rms)(fundamental) = V/Z√2 = (292.85)/(1.414 x 10.345) = 20.02 A
Fundamental Load power = (20.02)
2 x R = (20.02)
2 x 4 = 1603.2 Watts.
See lessA single phase full bridge inverter has RLC load with R = 4 Ω, L = 35 mH and C = 155 μF. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the rms value of the fundamental load current.
b Explanation: Average value of fundamental output voltage = 4Vs/π = 292.85 V X L = 2 x 3.14 x 50 x 0.035 = 10.99 Ω X C = 1/(2 x 3.14 x 50 x 155 x 10 -6 ) = 20.54 Ω Z = 10.345 Ω I = V/Z = 28.31 I(rms) = V/Z√2 = (292.85)/(1.414 x 10.345) = 20.02 A.
b
Explanation: Average value of fundamental output voltage = 4Vs/π = 292.85 V
X
L = 2 x 3.14 x 50 x 0.035 = 10.99 Ω
X
C = 1/(2 x 3.14 x 50 x 155 x 10
-6
) = 20.54 Ω
Z = 10.345 Ω
I = V/Z = 28.31
I(rms) = V/Z√2 = (292.85)/(1.414 x 10.345) = 20.02 A.
See lessA single phase full bridge inverter has RLC load. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the expression for the load voltage up to the fifth harmonic
b Explanation: In a single phase full bridge inverter only odd harmonics are present. i.e. 1,3,5 etc. Vo = (4Vs/π) sin ωt + (4Vs/3π) sin 3ωt + (4Vs/5π) sin 5ωt (4Vs/π) = 292 V ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t.
b
Explanation: In a single phase full bridge inverter only odd harmonics are present. i.e.
1,3,5 etc.
Vo = (4Vs/π) sin ωt + (4Vs/3π) sin 3ωt + (4Vs/5π) sin 5ωt
(4Vs/π) = 292 V
ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t.
See lessFor a full bridge inverter with the following load: R = 2 Ω, X L = 8 Ω and X C = 6 Ω
b Explanation: As the inductive effect is more than the capacitive effect, of course the current will lag the voltage by an angle P. P = tan -1 (X L – X C )/R = tan -1 (1) = 45°
b
Explanation: As the inductive effect is more than the capacitive effect, of course the
current will lag the voltage by an angle P.
P = tan
-1
(X
L – X
C
)/R = tan
-1
(1) = 45°
See lessA single phase full bridge inverter is fed from a dc source such that the fundamental component of output voltage = 230 V. Find the rms value of SCR and diode current respectively, for a R load of 2 Ω
d Explanation: Fundamental component of load current = V/R = 230/2 = 115 A. SCR current = 115/2 = 81.33 A Diode current = 0 as the diodes do not come into picture for R loads
d
Explanation: Fundamental component of load current = V/R = 230/2 = 115 A.
SCR current = 115/2 = 81.33 A
Diode current = 0 as the diodes do not come into picture for R loads
See lessA single phase full bridge inverter circuit, has load R = 2 Ω and dc source Vs = 230 V. Find the value of power delivered to the load in watts only due to the fundamental component of the load current
c Explanation: The fundamental component of voltage = (4Vs/π) sinωt Peak value Vm = 4Vs/π Rms voltage = 4Vs/π√2 = 207 V RMS Current (Irms) = 207/2 = 103.5 A P = (Irms) 2 x R = 21424.5 W.
c
Explanation: The fundamental component of voltage = (4Vs/π) sinωt
Peak value Vm = 4Vs/π
Rms voltage = 4Vs/π√2 = 207 V
RMS Current (Irms) = 207/2 = 103.5 A
P = (Irms)
2 x R = 21424.5 W.
See lessIn a half wave circuit, forced commutation is essential when the
b Explanation: When the load is resistive (R load) , the diodes do not conduct, hence they cannot help stop the conduction of the SCRs. Hence, forced commutation in such cases becomes essential.
b
Explanation: When the load is resistive (R load) , the diodes do not conduct, hence they
cannot help stop the conduction of the SCRs. Hence, forced commutation in such cases
becomes essential.
See lessIn a half wave bridge inverter circuit, the power delivered to the load by each source is given by
b Explanation: Power delivered by each source (Vs/2) each is (Vs/2) x Is
b
Explanation: Power delivered by each source (Vs/2) each is (Vs/2) x Is
See less