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The single-phase half-bridge inverter has a resistive load of 10Ω and the centre-tap dc input voltage is 96 V. The fundamental power consumed by the load is
- a)628.6 Watt
- b)525.0Watt'
- c)746.5 Watt
- d)824.4 Watt
Correct answer is option 'C'. Can you explain this answer?
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The single-phase half-bridge inverter has a resistive load of 10&Omega...
The nth harmonic-component of output voltage is
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The single-phase half-bridge inverter has a resistive load of 10&Omega...
A single phase half bridge inverter has a resistive load of R =10 and the dc input voltage vdc = 80v determine
rms value of output voltage
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rms value of output voltage
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The single-phase half-bridge inverter has a resistive load of 10&Omega...
The single-phase half-bridge inverter with a resistive load of 10 ohms operates by switching two power electronic devices, typically transistors, to alternate the direction of current flow through the load. This creates an AC output voltage waveform.
The operation of the half-bridge inverter can be divided into two modes: the positive half-cycle and the negative half-cycle.
During the positive half-cycle, one of the transistors is turned on, allowing current to flow from the DC input source through the load in one direction. The other transistor is turned off to prevent current flow in the opposite direction. This creates a positive half-cycle of the AC output voltage.
During the negative half-cycle, the state of the transistors is reversed. The previously turned-on transistor is turned off, while the previously turned-off transistor is turned on. This allows current to flow through the load in the opposite direction, creating a negative half-cycle of the AC output voltage.
By switching between these two modes at a high frequency, typically in the range of tens or hundreds of kilohertz, the half-bridge inverter can effectively generate an AC output voltage waveform.
The resistive load of 10 ohms determines the amount of current flowing through the load and thus the power dissipated. Using Ohm's Law (V = I * R), with a known load resistance of 10 ohms, the current flowing through the load can be calculated for a given output voltage.
For example, if the AC output voltage is 100 volts, the current flowing through the load would be 100 volts / 10 ohms = 10 amps. The power dissipated in the load can then be calculated using the formula P = I^2 * R, which would be 10 amps^2 * 10 ohms = 1000 watts or 1 kilowatt.
The operation of the half-bridge inverter can be divided into two modes: the positive half-cycle and the negative half-cycle.
During the positive half-cycle, one of the transistors is turned on, allowing current to flow from the DC input source through the load in one direction. The other transistor is turned off to prevent current flow in the opposite direction. This creates a positive half-cycle of the AC output voltage.
During the negative half-cycle, the state of the transistors is reversed. The previously turned-on transistor is turned off, while the previously turned-off transistor is turned on. This allows current to flow through the load in the opposite direction, creating a negative half-cycle of the AC output voltage.
By switching between these two modes at a high frequency, typically in the range of tens or hundreds of kilohertz, the half-bridge inverter can effectively generate an AC output voltage waveform.
The resistive load of 10 ohms determines the amount of current flowing through the load and thus the power dissipated. Using Ohm's Law (V = I * R), with a known load resistance of 10 ohms, the current flowing through the load can be calculated for a given output voltage.
For example, if the AC output voltage is 100 volts, the current flowing through the load would be 100 volts / 10 ohms = 10 amps. The power dissipated in the load can then be calculated using the formula P = I^2 * R, which would be 10 amps^2 * 10 ohms = 1000 watts or 1 kilowatt.
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The single-phase half-bridge inverter has a resistive load of 10Ω and the centre-tap dc input voltage is 96 V. The fundamental power consumed by the load isa)628.6 Wattb)525.0Watt'c)746.5 Wattd)824.4 WattCorrect answer is option 'C'. Can you explain this answer?
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