A typical incandescent bulb has an efficiency of about 10%. What current is needed for a bulb with a resistance of 1.5×10^2 Ω to output power of 6.0 W?
Explanations & Calculations
P=6.0 W×10010=60.0 W\qquad\qquad \begin{aligned} \small P&=\small 6.0\,W\times\frac{100}{10}=60.0\,W \end{aligned}P=6.0W×10100=60.0W
P=I2RI=60.0 W1.5×102 Ω=6.3×10−1 A\qquad\qquad \begin{aligned} \small P&=\small I^2R\\ \small I &=\small \sqrt{\frac{60.0\,W}{1.5\times10^2\, \Omega}}= 6.3\times10^{-1}\,A \end{aligned}PI=I2R=1.5×102Ω60.0W=6.3×10−1A
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