A single phase full bridge inverter feeds power at 50 Hz to RLC load R - 5 - L - 0-3 H and C- 50 -F- The dc input voltage is 220 V- The power absorbed by the load isWatts-

X_L=2π fL=30πΩ X_C=12π fC=63.66 Ω X=(X_L X_C)=(30π 63.66)Ω=30.58 Ω Z⃗_⃗1⃗=R+jX=5+j30.58=30.98 ∠80.71^o Output voltage expression of 1 ϕ FBI can be given ...

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Question

A single phase full bridge inverter feeds power at 50 Hz to RLC load R = 5 $Ω$ , L = 0.3 H and C= 50 $μ$ F. The dc input voltage is 220 V. The power absorbed by the load is______Watts.

102.14

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204.28

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250

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198

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Solution

The correct option is B 204.28
$X_{L} = 2 π f L = 30 π Ω$

$X_{C} = \frac{1}{2 π f C} = 63.66 Ω$

$X = (X_{L} - X_{C}) = (30 π - 63.66) Ω = 30.58 Ω$

$_{1} = R + j X = 5 + j 30.58 = 30.98 ∠ {80.71}^{o}$

Output voltage expression of 1 - $ϕ$ FBI can be given as

$V_{o n} = \infty \sum n = 1, 3, 5.. \frac{4 V_{d c}}{n π} sin n ω_{o} t$

$V_{01} = \frac{2 \sqrt{2} V_{d c}}{π} \times sin (314 t) = 198 \sqrt{2} sin 314 t V$

$i_{01} = 6.3912 \sqrt{2} sin (314 t - 80.71) A$

$P_{o} = V_{1 r} I_{1 r} cos (ϕ_{1})$

$= 198 \times 6.3912 cos (80.71) = 204.28 W a t t$

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A single phase full bridge inverter feeds power at 50 Hz to RLC load R = 5 Ω, L = 0.3 H and C= 50 μF. The dc input voltage is 220 V. The power absorbed by the load is______Watts.

A single phase full bridge inverter feeds power at 50 Hz to RLC load R = 5 $Ω$ , L = 0.3 H and C= 50 $μ$ F. The dc input voltage is 220 V. The power absorbed by the load is______Watts.