An electric bulb is designed to draw `P_(0)` power at `V_(0)` voltage. If the voltage is `V`, it drawas power. ThenA. `P=((V_(0))/(V))^(2)P_(0)`B. `P=((V)/(V_(0)))^(2)P_(0)`C. `P=((V)/(V_(0)))P_(0)`D. `P=((V_(0))/(V))P_(0)`

An electric bulb is designed to draw `P_(0)` power at `V_(0)` voltage.

1.	An electric bulb is designed to draw `P_(0)` power at `V_(0)` voltage. If the voltage is `V`, it drawas power. ThenA. `P=((V_(0))/(V))^(2)P_(0)`B. `P=((V)/(V_(0)))^(2)P_(0)`C. `P=((V)/(V_(0)))P_(0)`D. `P=((V_(0))/(V))P_(0)`
Answer» Correct Answer - B

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An electric bulb is designed to draw `P_(0)` power at `V_(0)` voltage. If the voltage is `V`, it drawas power. ThenA. `P=((V_(0))/(V))^(2)P_(0)`B. `P=((V)/(V_(0)))^(2)P_(0)`C. `P=((V)/(V_(0)))P_(0)`D. `P=((V_(0))/(V))P_(0)`

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