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The voltage required across the lamp is 100V and at this voltage the current through both bulb and capacitor will be I=0.5 A to dissipate 50W in the bulb.

First we need the voltages across bulb and capacitor to combine to give 200V. With the current through each being in phase then the voltage V across the capacitor will be 90° out of phase with that across the bulb (which I’m assuming is incandescent and therefore purely resistive). So we need to use Pythagoras to combine them:

1002+V2=2002

and so V=1003–√ .

Now the relationship between AC voltage and current for a capacitor is

I=2πfCV

where frequency f is 50Hz in this case. Solving for C and inserting the values for I and V calculated above we get:

C=I2πfV

=0.510000π3–√

=50π3–√μF

This is approximately 9.19μ F .

Answer:

there were 2 Chandragupta

Answer:

First of all, covert all the given time in terms of second.

1 minute = 60 seconds

2 minutes 24 seconds = 144 seconds

1 hour = (60×60) seconds = 3600 seconds

Required % = 144×100/3600

= 4 %

Answer:

The percentage is 10 minutes 43 seconds of an hour is 17.86 %

Step-by-step explanation:

• A percentage is the ratio of any value to the whole value multiplied by 100.
• Number of seconds in 1 min =60
• Number of seconds in 1 hour=60 ×60 =3600

• Given minutes is 10 min 43 seconds
• Covert the given time in terms of the second.

      10 min 43 seconds  =(10x60)+43

                                        [tex]=643[/tex]

• Percentage of the time is given by

         [tex]\%= \frac{643}{3600}[/tex] ×100

        [tex]\\\%=17.86 \%[/tex]

The percentage is 10 minutes 43 seconds of an hour is 17.86 %

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Here is ur solution

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