A sawtooth waveform is fed to "Average reading (with full scale rectification)" ac electronic voltmeter. This voltmeter is calibrated for RMS value of pure sinusoidal input. What will be reading displayed on voltmeter? Also find percentage error in the reading. Take amplitude as 10 volts and time period of 1 second.

The given answers are:

1. RMS indication = 6.5 V
2. error = 12.7%

#Here's what I have done so far

\$V_{RMS}\$ of sawtooth is 5.77 V and average is 5 V. \$V_{RMS}\$ of sinusoid is 7.07 V and average is 6.37 V. So the ratio of \$V_{RMS}\$ to \$V_{avg}\$ is 1.11 and 1.154 of sinusoid and sawtooth respectively. But the question here is that voltage of sawtooth waveform is being measured by sinusoidally calibrated voltmeter and I don't understand how to find the value that will be measured and what will be the error.

A sawtooth waveform is fed to "Average reading (with full scale rectification)" ac electronic voltmeter. This voltmeter is calibrated for RMS value of pure sinusoidal input. What will be reading displayed on voltmeter? Also find percentage error in the reading. Take amplitude as 10 volts and time period of 1 second.

The given answers are:

1. RMS indication = 6.5 V
2. error = 12.7%

Here's what I have done so far

\$V_{RMS}\$ of sawtooth is 5.77 V and average is 5 V. \$V_{RMS}\$ of sinusoid is 7.07 V and average is 6.37 V. So the ratio of \$V_{RMS}\$ to \$V_{avg}\$ is 1.11 and 1.154 of sinusoid and sawtooth respectively. But the question here is that voltage of sawtooth waveform is being measured by sinusoidally calibrated voltmeter and I don't understand how to find the value that will be measured and what will be the error.

A sawtooth waveform is fed to "Average reading (with full scale rectification)" ac electronic voltmeter. This voltmeter is calibrated for RMS value of pure sinusoidal input. What will be reading displayed on voltmeter? Also find percentage error in the reading. Take amplitude as 10 volts and time period of 1 second."

Please help me. I am not able to solve it since 1 hour. No idea what to do. Answers are:

1. RMS indication - 6.5 V
2. error - 12.7%

I did like this - rms of sawtooth is 5.77 V. avg is 5 V. rms of sinusoid is 7.07 V and average is 6.37 V.Form factor is 1.11 and 1.154 of sinusoid and sawtooth respectively But the question here is that voltage of sawtooth waveform is being measured by sinusoidally calibrated voltmeter .So what will be the value that will be measured and waht will be the error?

A sawtooth waveform is fed to "Average reading (with full scale rectification)" ac electronic voltmeter. This voltmeter is calibrated for RMS value of pure sinusoidal input. What will be reading displayed on voltmeter? Also find percentage error in the reading. Take amplitude as 10 volts and time period of 1 second.

The given answers are:

1. RMS indication = 6.5 V
2. error = 12.7%

#Here's what I have done so far

\$V_{RMS}\$ of sawtooth is 5.77 V and average is 5 V. \$V_{RMS}\$ of sinusoid is 7.07 V and average is 6.37 V. So the ratio of \$V_{RMS}\$ to \$V_{avg}\$ is 1.11 and 1.154 of sinusoid and sawtooth respectively. But the question here is that voltage of sawtooth waveform is being measured by sinusoidally calibrated voltmeter and I don't understand how to find the value that will be measured and what will be the error.