- #1
Manelectro
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I'm having a tough time wrapping my head around something that seems so simple. The Question is:
What is the maximum output power from a 12 volt battery that has an internal resistance of 6 ohm ?
The answer from this books is 6 W?
My formula
First I need to find the current
I = V/R so 12volt/6 ohm = 2 ampere
When is use this formula
P = VI, 12Vx2A = 24Watt,
The next question is:
On an oscilloscope display of a pure sine wave , the peak to peak value is 7 volts. The average value of the display voltage is about:
A. 6.36 volt
B.3.18 volt
C. 0 volt
D. 2.23 volt
The answer from this book is 0 volt.
The peak to peak voltage is 7/2 = 3.5V this peak voltage
But by calculation Vaverage = 2 x 3.5/ pi = 2.228 or 2.23 V for sine wave or AC wave formula.
If the average value were taken over the whole cycle then the result would be zero value.
Thank you,
Azman
What is the maximum output power from a 12 volt battery that has an internal resistance of 6 ohm ?
The answer from this books is 6 W?
My formula
First I need to find the current
I = V/R so 12volt/6 ohm = 2 ampere
When is use this formula
P = VI, 12Vx2A = 24Watt,
The next question is:
On an oscilloscope display of a pure sine wave , the peak to peak value is 7 volts. The average value of the display voltage is about:
A. 6.36 volt
B.3.18 volt
C. 0 volt
D. 2.23 volt
The answer from this book is 0 volt.
The peak to peak voltage is 7/2 = 3.5V this peak voltage
But by calculation Vaverage = 2 x 3.5/ pi = 2.228 or 2.23 V for sine wave or AC wave formula.
If the average value were taken over the whole cycle then the result would be zero value.
Thank you,
Azman