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P Pankaj Sanodiya
Power required  The total resistance of the two-wire line Input Voltage  Output Voltage: RMS current in the wireline Now, a) power loss in the line b) Power supplied by plant = 800 kW + 6 kW = 806kW. c) Voltage drop in the power line =  Total voltage transmitted from the plant = 300+40000=40300 as power is generated at 440V, The rating of the power plant is 440V-40300V. We prefer high...

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P Pankaj Sanodiya
Power required  The total resistance of the two-wire line Input Voltage  Output Voltage: RMS Current in the wireline Now, Voltage drop in the power line =  Total voltage transmitted from the plant = 3000+4000=7000 as power is generated at 440V, The rating of the power plant is 440V-7000V.

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P Pankaj Sanodiya
Given, Input voltage:   Number of turns in the primary coil Output voltage: Now, Let number of turns in secondary be    Now as we know, in a transformer, Hence the number of turns in secondary winding id 400.    

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P Pankaj Sanodiya
We need choke coil in the use of fluorescent tubes with ac mains to reduce the voltage across the tube without wasting much power. If we use simply resistor for this purpose, there will be more power loss, hence we do not prefer it.

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P Pankaj Sanodiya
For a steady state DC, the increasing inductance value by inserting iron core in the choke, have no effect in the brightness of the connected lamp, whereas, for ac when the iron core is inserted, the light of the lamp will shine less brightly because of increase in inductive impedance.

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P Pankaj Sanodiya
For a high frequency, the inductive reactance and capacitive reactance:   Hence the capacitor does not offer resistance to a higher frequency, so the ac voltage appears across L. Similarly For DC, the inductive reactance and capacitive reactance: Hence DC signal appears across Capacitor only.

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P Pankaj Sanodiya
Yes, we use capacitors in the primary circuit of an induction coil to avoid sparking. when the circuit breaks, a large emf is induced and the capacitor gets charged from this avoiding the case of sparking and short circuit.

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P Pankaj Sanodiya
Yes, at any instant the applied voltage will be distributed among all element and the sum of the instantaneous voltage of all elements will be equal to the applied. But this is not the case in RMS because all elements are varying differently and they may not be in the phase.

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P Pankaj Sanodiya
The inductance of the inductor  The capacitance of the capacitor  The resistance of the resistor  Now, Resonant frequency Q-Factor of the circuit Now, to improve the sharpness of resonance by reducing its full width at half maximum, by a factor of 2 without changing, we have to change the resistance of the resistor to half of its value, that is

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P Pankaj Sanodiya
As Power  Power  will be half when the current  is   times the maximum current. As,    At half power point : here, On putting values, we get, two values of  for which   And they are: Also, The current amplitude at these frequencies

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P Pankaj Sanodiya
The value of maximum angular frequency is calculated in the first part of the question and whose magnitude is 4166.67 Q-factor of any circuit is given by Hence Q-factor for the circuit is 21.74.

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P Pankaj Sanodiya
Since the resistor is the only element in the circuit which consumes the power, the maximum absorbed power by circuit will be maximum when power absorbed by the resistor will be maximum. power absorbed by the resistor will be maximum at when current is maximum which is the natural frequency case, Hence when source frequency will be equal to the natural frequency, the power absorbed will be...

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P Pankaj Sanodiya
The inductance of the inductor  The capacitance of the capacitor  The resistance of the resistor  Voltage supply  Frequency of voltage supply  As we know, the current amplitude is maximum at the natural frequency of oscillation, which is  Also, at this frequency, SO, The maximum current in the circuit : Hence maximum current is 14.14A.

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P Pankaj Sanodiya
The inductance of the inductor  The capacitance of the capacitor  The resistance of a resistor  Voltage supply  Frequency of voltage supply  As we know, Impedance   Current flowing in the circuit : Now, Average power transferred to the resistor: Average power transferred to the inductor = 0 Average power transferred to the capacitor = 0: Total power absorbed by circuit : Hence circuit...

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P Pankaj Sanodiya
Since the phase difference between voltage and current is 90 degree, even the total power absorbed by the circuit is zero. This is an ideal circuit, we can not have any circuit in practical that consumes no power, that is because practically resistance of any circuit is never zero. Here only inductor and capacitor are present and none of them consumes energy, they just store it and transfer it...

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P Pankaj Sanodiya
As we know, Average power   where  is the phase difference between voltage and current. Since in the circuit, phase difference  is , the average power is zero.

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P Pankaj Sanodiya
Since   Current flowing in the circuit is sinusoidal and hence average power will be zero as the average of sin function is zero.in other words, the inductor will store energy in the positive half cycle of the sin (0 degrees to 180 degrees) and will release that energy in the negative half cycle(180 degrees to 360 degrees), and hence average power is zero.

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P Pankaj Sanodiya
As we know, RMS potential drop across an element with impedance Z: SO, RMS potential difference across inductor: RMS potential drop across capacitor
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