HW4solutions - Physics 213 HW #4 Solutions Spring 2009...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 213 HW #4 – Solutions Spring 2009 Q25.8 [Batteries] Labeling batteries by emf makes sense because the potential difference across the battery is constant in the ideal case, and nearly constant for real batteries. Labeling batteries by current would not be appropriate, because the current depends on the resistance across the circuit that you hook the battery up to. (By Ohm’s law, I=V/R.) 25.45. [European Light Bulb] A “100-W” European bulb dissipates 100 W when used across 220 V. (a) The voltage applied to the light bulb will be different in America, but the resistance R will be the same since R only depends on the light bulb itself. Then we can take the ratio of the power in the US to the power in Europe, using P = V 2 / R . 2 2 2 US US US 2 E E E / 120 V . / 220 V P V R V P V R V    This gives 2 US 120 V (100 W) 220 V P = 29.8 W. The bulb draws considerably less power in the U.S., so it would be much dimmer than in Europe. (b) Use P = IV to find the current. I = P/V = (29.8 W)/(120 V) = 0.248 A 25.86. [Power Match] The short-circuit current is the current we’d get if we connect the ends of the battery with a zero resistance wire. The short- circuit current is short circuit /. Ir Note that, if the battery had no internal resistance, the short-circuit current would be infinite! The power output of the source is ( ) . VI Ir I  (a) 2 P I I r , so 20 dP Ir dI for maximum power output and max short circuit 11 . 22 P II r  (b) Applying Ohm’s law with two resistors (R and r) in series gives effective I R r R . Setting this current to the max P I calculated in part (a) gives 1 2 I r R r  . Therefore, 2 r R r  and . Rr As before, the power output of the source is 2 / P VI I R . Then, 2 2 2 . 24 P I R r rr When R is smaller than r , I is large and the 2 losses in the battery are large. When R is larger than r , I is small and the power output I of the battery emf is small. Q26.4 [Bulb Brightnesses] Bulbs B and C are in series. Bulb A is in parallel with B and C. The bulbs have identical resistance R. The effective resistance across the path which goes through A is half the resistance of the path that goes across B and C. Since there is the same potential difference across both paths, the current through A must be twice the current through B and C, so A will be brightest. A also has the largest potential difference (because V=IR, and because the same potential difference covered by B+C is covered by A alone). B and C are in series, so the same current passes through them, and they have the same brightness. If we unscrew bulb A, then A will become an open switch and current will not flow through the A path.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/14/2009 for the course PHYS 2213 taught by Professor Perelstein,m during the Spring '07 term at Cornell University (Engineering School).

Page1 / 5

HW4solutions - Physics 213 HW #4 Solutions Spring 2009...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online