Question:
A single-phase half-controlled bridge converter supplies an inductive load with ripple-free load current. The triggering angle of the converter is \( 60^\circ \). The ratio of the rms value of the fundamental component of the input current to the rms value of the total input current of the bridge is (rounded off to 3 decimal places).
A single-phase half-controlled bridge converter supplies an inductive load with ripple-free load current. The triggering angle of the converter is \( 60^\circ \). The ratio of the rms value of the fundamental component of the input current to the rms value of the total input current of the bridge is (rounded off to 3 decimal places).
Show Hint
For bridge converters, use Fourier analysis to separate fundamental and harmonic components accurately.
Updated On: Jan 23, 2025
Hide Solution
Verified By Collegedunia
Solution and Explanation
Step 1: Calculate the fundamental rms component.
Using Fourier series analysis, determine the rms value of the fundamental current component.
Step 2: Calculate the total rms current.
The total rms current includes all harmonics. Use the formula for the ratio:
\[
\text{Ratio} = \frac{\text{Fundamental rms component}}{\text{Total rms current}}.
\]
After calculations:
\[
\text{Ratio} = 0.940 \, \text{to} \, 0.970.
\]
Was this answer helpful?
0
0
Top Questions on Current
- In the DC-DC converter shown in the figure, the current through the inductor is continuous. The switching frequency is \( 500 \, \text{Hz} \). The voltage (\( V_o \)) across the load is assumed to be constant and ripple-free. The peak inductor current in amperes is (rounded off to the nearest integer).
\includegraphics[width=0.5\linewidth]{64.png} - A single-phase full-controlled thyristor converter bridge is used for regenerative braking of a separately excited DC motor with the following specifications: Rated armature voltage = \( 210 \, \text{V} \), Rated armature current = \( 10 \, \text{A} \), Rated speed = \( 1200 \, \text{rpm} \), Armature resistance = \( 1 \, \Omega \), Input to the converter bridge = \( 240 \, \text{V at 50 Hz} \). Assume that the motor is running at \( 600 \, \text{rpm} \) and the armature terminals of the motor are suitably reversed for regenerative braking. If the armature current of the motor is to be maintained at the rated value, the triggering angle of the converter bridge in degrees should be (rounded off to 2 decimal places).
\begin{tabular}{|l|l|} \hline Parameter & Value
\hline Rated armature voltage & 210 V
\hline Rated armature current & 10 A
\hline Rated speed & 1200 rpm
\hline Armature resistance & 1 $\Omega$
\hline Input to the converter bridge & 240 V at 50 Hz
\hline \end{tabular}
The armature of the DC motor is fed from the fullcontrolled bridge and the field current is kept constant. - A 5 kW, 220 V DC shunt motor has \( 0.5 \, \Omega \) armature resistance including brushes. The motor draws a no-load current of \( 3 \, A \). The field current is constant at \( 1 \, A \). Assuming that the core and rotational losses are constant and independent of the load, the current (in amperes) drawn by the motor while delivering the rated load, for the best possible efficiency, is (rounded off to 2 decimal places).
- Consider the closed-loop system shown in the figure with \[ G(s) = \frac{K(s^2 - 2s + 2)}{s^2 + 2s + 5}. \] The root locus for the closed-loop system is to be drawn for \( 0 \leq KLt;\infty \). The angle of departure (between \( 0^\circ \) and \( 360^\circ \)) of the root locus branch drawn from the pole \( (-1 + j2) \), in degrees, is (rounded off to the nearest integer).
- Consider the stable closed-loop system shown in the figure. The asymptotic Bode magnitude plot of \( G(s) \) has a constant slope of \( -20 \, \text{dB/decade} \) at least till \( 100 \, \text{rad/sec} \) with the gain crossover frequency being \( 10 \, \text{rad/sec} \). The asymptotic Bode phase plot remains constant at \( -90^\circ \) at least till \( \omega = 10 \, \text{rad/sec} \). The steady-state error of the closed-loop system for a unit ramp input is (rounded off to 2 decimal places).
\includegraphics[width=0.5\linewidth]{58.png}
View More Questions
Questions Asked in GATE EE exam
- A single-phase full bridge voltage source inverter (VSI) feeds a purely inductive load. The inverter output voltage is a square wave in \( 180^\circ \) conduction mode. The fundamental frequency of the output voltage is \( 50 \, \text{Hz} \). If the DC input voltage of the inverter is \( 100 \, \text{V} \) and the value of the load inductance is \( 20 \, \text{mH} \), the peak-to-peak load current in amperes is (rounded off to the nearest integer).
- A 3-phase star connected slip ring induction motor has the following parameters referred to the stator: \[ R_s = 3 \, \Omega, \, X_s = 2 \, \Omega, \, X_r' = 2 \, \Omega, \, R_r' = 2.5 \, \Omega \] The per phase stator to rotor effective turns ratio is 3:1. The rotor winding is also star connected. The magnetizing reactance and core loss of the motor can be neglected. To have maximum torque at starting, the value of the extra resistance in ohms (referred to the rotor side) to be connected in series with each phase of the rotor winding is \_\_\_\_\_\_ (rounded off to 2 decimal places).
- GATE EE - 2024
- Voltage and Frequency control
- The single line diagram of a lossless system is shown in the figure. The system is operating in steady-state at a stable equilibrium point with the power output of the generator being \( P_{max} \sin \delta \), where \( \delta \) is the load angle and the mechanical power input is \( 0.5 P_{max} \). A fault occurs on line 2 such that the power output of the generator is less than \( 0.5 P_{max} \) during the fault. After the fault is cleared by opening line 2, the power output of the generator is \( \frac{P_{max}}{\sqrt{2}} \sin \delta \). If the critical fault clearing angle is \( \pi/2 \) radians, the accelerating area on the power angle curve is \_\_\_ times \( P_{max} \) (rounded off to 2 decimal places).
\includegraphics[width=0.5\linewidth]{56.png} - Two passive two-port networks \( P \) and \( Q \) are connected as shown in the figure. The impedance matrix of network \( P \) is \( Z_P = \begin{bmatrix} 40 \, \Omega & 60 \, \Omega
80 \, \Omega & 100 \, \Omega \end{bmatrix} \). The admittance matrix of network \( Q \) is \( Y_Q = \begin{bmatrix} 5 \, S & -2.5 \, S
-2.5 \, S & 1 \, S \end{bmatrix} \). Let the ABCD matrix of the two-port network \( R \) in the figure be \( \begin{bmatrix} \alpha & \beta
\gamma & \delta \end{bmatrix} \). The value of \( \beta \) in \( \Omega \) is \_\_\_\_\_\_ (rounded off to 2 decimal places).
\includegraphics[width=0.5\linewidth]{48.png}- GATE EE - 2024
- Miscellaneous
- In the circuit shown, \( Z_1 = 50 \angle -90^\circ \, \Omega \) and \( Z_2 = 200 \angle -30^\circ \, \Omega \). It is supplied by a three-phase 400 V source with the phase sequence being R-Y-B. Assume the wattmeters \( W_1 \) and \( W_2 \) to be ideal. The magnitude of the difference between the readings of \( W_1 \) and \( W_2 \) in watts is \_\_\_\_\_\_ (rounded off to 2 decimal places).
\includegraphics[width=0.5\linewidth]{50.png}- GATE EE - 2024
- Miscellaneous
View More Questions