Voltage of an AC source is given by V- 200sin100- t-cos100- t where t is in seconds and V is in volts- Then the peak voltage isvolts-

V=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

Voltage of an...

Question

Voltage of an AC source is given by 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 is
volts.

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100

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200

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400

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Solution

The correct option is B 100
$V = 100 \times 2 s i n 100 π t c o s 100 π t = 100 s i n 200 π t$
$\Rightarrow V_{0} = 100$ Volts and Frequency =100Hz

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Voltage of an AC source is given by V= 200sin(100π t).cos(100π t) where t is in seconds and V is in volts. Then the peak voltage is volts.

The correct option is B 100V=100× 2sin 100π t cos 100 π t=100 sin 200π t ⇒ V0=100 Volts and Frequency =100Hz

Voltage of an AC source is given by 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 is
volts.

The correct option is B 100
$V = 100 \times 2 s i n 100 π t c o s 100 π t = 100 s i n 200 π t$
$\Rightarrow V_{0} = 100$ Volts and Frequency =100Hz