spring_05_exam2_sol_phy2049

spring_05_exam2_sol_phy2049 - PHY2049 Spring 2005 Prof....

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
PHY2049 Spring 2005 1 Prof. Darin Acosta Prof. Paul Avery March 9, 2005 PHY2049, Spring 2005 Exam 2 Solutions 1. Two identical light bulbs A and B are connected in series to a constant voltage source. A wire is then connected across B. What is the brightness of A relative to its former brightness? A light bulb will have a certain resistance R. The light it emits will be proportional to the energy dissipated by this resistance: 2 PiR = . So when two identical light bulbs are connected in series to a constant voltage V, the current passing through the circuit is /2 iV R = and the power dissipated in light bulb A is: 22 1 2 44 VV PR R R == . Now when a wire is connected across light bulb B, that means that the resistance is bypassed by a wire in parallel to the bulb (i.e. the bulb is shorted out by a jumper). The resistance of bulb B becomes zero, and the current in the circuit is / iVR = . This current is twice as high as before, but the power dissipated by light bulb A will be four times more: 2 21 4 V PP R Answer: 4 2. In the circuit shown, what is the current (in amps) flowing through the 18V battery? Let 1 i be the current through the 18V battery, 2 i the current through the middle branch with the 6V battery, and 3 i the current through the rightmost branch. By Kirchoff’s junction rule: 123 iii =+ . By Kirchoff’s voltage sum rule applied to the rightmost loop: 23 32 6V 12 24 0.5 0.25 ii +Ω= Ω ⇒= +
Background image of page 1

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

View Full DocumentRight Arrow Icon
PHY2049 Spring 2005 2 By Kirchoff’s voltage sum rule applied to the leftmost loop: () 12 23 2 22 2 2 32 123 18V 6 6V 12 18 6 6 12 18 6 6 0.25 0.5 6 12 18 1.5 9 6 12 10.5 0.5 21 0.25 0.5 0.5 1.0 ii i i i iii −Ω= + Ω −+ = + −− + = + −−= + ⇒= = + = ⇒=+= Answer: 1.0 Amps through 18V battery 3. An electron is accelerated from rest by a potential difference of 8.0 kV. It then enters a uniform magnetic field of magnitude 1.0 T with its velocity perpendicular to the direction
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

spring_05_exam2_sol_phy2049 - PHY2049 Spring 2005 Prof....

This preview shows document pages 1 - 3. Sign up to view the full document.

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