The question with its answer is as shown below:
However, I have solved it as follows:
Am I right or not?
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\$\begingroup\$ Omar, are you done with this question now? Do you feel ready to accept the answer or, do you need clarification? \$\endgroup\$– Andy akaCommented Jan 8, 2021 at 12:58
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1\$\begingroup\$ Done. Thanks. Am I supposed to reply Done? \$\endgroup\$– OMARCommented Jan 8, 2021 at 13:06
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1\$\begingroup\$ Thanks I honestly forgot. \$\endgroup\$– OMARCommented Jan 8, 2021 at 13:51
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1 Answer
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What is the amplitude of the output voltage?
Your answer is fine but you have misinterpreted what is meant by output amplitude. Whenever we talk about a signal amplitude (with no other constraints) we talk about an RMS value hence, you calculated the RMS value correctly at 20 amps but then assumed that the peak value of 28.28 volts was required. It isn't required.
An alternative way recognizes that the 200 μF capacitor has an impedance magnitude of 50 Ω when ω = 100 radians per second. This means that the magnitude of Vout/Vin is 0.7071. Input RMS is 40 x 0.7071 therefore output amplitude is 20 volts.
