An electric bulb is designed to draw power P 0 at voltage V 0 . If the voltage is V and…
Physics · JEE Main · NTA Exams — Current Electricity
An electric bulb is designed to draw power P0 at voltage V0. If the voltage is V and it draws a power P. Then:
- \(p^{2}=\left(\frac{V_{B}}{V}\right)^{2} p_{B}\)
- \(P^{2}=\left(\frac{V}{V_{B}}\right)^{2} P_{B}\)
- \(P=\left(\frac{V}{V_{B}}\right) P_{B}\)
- \(P^{2}=\left(\frac{V_{B}}{V}\right) P_{B}\)
Answer
(B) P^ 2 = ( V V_ B )^ 2 P_ B
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