The step-down transformer has a ratio of 10:1

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$$P_{globe}=V_{globe}\cdot I_{globe}$$ $$P_{globe}=V_{globe}\cdot I_{globe}=\left( 2 \right)\cdot I_{globe}=4\ \mbox{W}$$ $$I_{globe}=2\; \mbox{A}\; ∴\; I=2\cdot \frac{1}{10}=0.2\; \mbox{A}$$ $$V_{loss}=IR=\left( 0.2 \right)\left( 2+2 \right)=0.8\; \mbox{V}$$ $$However...$$ $$V_{loss}=V_{supply}-\frac{10}{1}V_{globe}=20.8\sqrt{2}-10\left( 2 \right)≈9.4\; \mbox{V}$$

What's wrong here?


The \$\sqrt 2\$ factor should be removed from Vsupply because you are comparing rms values and not peak values.

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  • \$\begingroup\$ also just a quick question, is 2P(peak)=P(rms) ? \$\endgroup\$ – inspd Nov 2 '15 at 11:25
  • \$\begingroup\$ The instantaneous power P(t)=V(t)·I(t). Therefore P(peak)=max{P(t)}. If V(t) is a sine, and I(t) is in phase with it as in this problem (considering an ideal transformer and resistive globe), we'll have P(peak)=\$\sqrt 2 V_{rms} \cdot \sqrt 2 I_{rms}=2P_{rms}\$. But usually no one uses P(peak), in an alternate current problem everything should be considered rms if not told the contrary. \$\endgroup\$ – Roger C. Nov 2 '15 at 11:41

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