1
\$\begingroup\$

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

enter image description here


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


What's wrong here?

\$\endgroup\$
2
\$\begingroup\$

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

\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.