The voltage of an ac source varies with time according to the equation V -200 sin 100 - t -cos 100 - t where t is in seconds and V is in volts- ThenA- The peak voltage of the source is 100 voltsB- The peak voltage of the source is 50 voltsC- The frequency of the source is 50 HzD- The peak voltage of the source is 100 - -2 volts

The correct option is A The peak voltage of the source is 100 voltsV=100× 2sin 100π t cos 100 π t=100 sin 200π t ⇒ V_0=100 Volts and Frequency =100Hz ...

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Standard XII

Physics

Sinusoidal AC

The voltage o...

Question

The voltage of an ac source varies with time according to the equation V = 200sin $(100 π t) . c o s (100 π t)$ where t is in seconds and V is in volts. Then

The peak voltage of the source is 100 volts

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The peak voltage of the source is 50 volts

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The peak voltage of the source is 100 / √2 volts

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The frequency of the source is 50 Hz

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Solution

The correct option is A The peak voltage of the source is 100 volts
$V = 100 \times 2 s i n 100 π t c o s 100 π t = 100 s i n 200 π t$
$\Rightarrow V_{0} = 100$ Volts and Frequency =100Hz

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The voltage of an ac source varies with time according to the equation V = 200sin(100π t).cos(100π t) where t is in seconds and V is in volts. Then

The correct option is A The peak voltage of the source is 100 voltsV=100× 2sin 100π t cos 100 π t=100 sin 200π t ⇒ V0=100 Volts and Frequency =100Hz

The voltage of an ac source varies with time according to the equation V = 200sin $(100 π t) . c o s (100 π t)$ where t is in seconds and V is in volts. Then

The correct option is A The peak voltage of the source is 100 volts
$V = 100 \times 2 s i n 100 π t c o s 100 π t = 100 s i n 200 π t$
$\Rightarrow V_{0} = 100$ Volts and Frequency =100Hz