HW5solutions-10 - Physics 2213 HW #5 Solutions Spring 2010...

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

View Full Document Right Arrow Icon
Physics 2213 HW #5 – Solutions Spring 2010 26.23. [Kirchoff’s Circuit Rules] We have to use Kirchoff’s rules to find the unknown currents, emfs, and resistances. Apply the junction rule at points a, b, c and d to calculate the unknown currents. Then apply the loop rule to three loops to calculate 12 , and . R εε The circuit is sketched to the right. (a) Apply the junction rule to point a : 3 3.00 A 5.00 A 0 I + −= 3 8.00 A I = Apply the junction rule to point b : 4 2.00 A 3.00 A 0 I +− = 4 1.00 A I = Apply the junction rule to point c : 345 0 III −−= 534 8.00 A 1.00 A 7.00 A =−= = As a check, apply the junction rule to point d : 5 2.00 A 5.00 A 0 I 5 7.00 A I = (b) Apply the loop rule to loop(1): ( )( ) ( ) 13 3.00 A 4.00 3.00 0 I ε Ω− Ω= ( )( ) 1 12.0 V 8.00 A 3.00 36.0 V = + Apply the loop rule to loop (2): ( )( ) ( ) 23 5.00 A 6.00 3.00 0 I ( )( ) 2 30.0 V 8.00 A 3.00 54.0 V = + (c) Apply the loop rule to loop (3): ( ) 2.00 A 0 R −+ = 21 54.0 V 36.0 V 9.00 2.00 A 2.00 A R −− = = = Finally, we can apply the loop rule to loop (4) as a check of our calculations: ( ) ( )( ) ( )( ) 2.00 A 3.00 A 4.00 5.00 A 6.00 0 R −Ω +Ω = ( )( ) 2.00 A 9.00 12.0 V 30.0 V 0 + = 18.0 V 18.0 V 0 −+= 26.50. [Electric Dryer] (a) VI P = , so I = P V = 4100 W 240 V = 17.1 A. Recall from page 201 that 12-gauge wire can carry up to 20A safely. Thus, we can use 12-gauge wire safely for this electric dryer. (b) 2 V P VI R = = . Solving for the resistance, R = V 2 P = (240 V) 2 4100 W = 14 . (c) At 11 / c per kWH, for 1 hour the cost is (11 / c/kWh)(1 h)(4.1 kW) = 45 / c . (d) It makes sense that the cost to operate the device is proportional to its power consumption. The number also sounds about right. I am charged about one dollar to run a load of clothes through a dryer for an hour.
Background image of page 1

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

View Full DocumentRight Arrow Icon
- 2 - 26.92 [Resistor Cube] Kirchoff’s junction rule: The current I entering the cube divides equally into three currents, I /3, since the resistance along each of the three branches is the same. When the current I /3 reaches the next junction, it again divides equally, this time into two currents of I /6. Finally, when the three currents reach the last junction, they must equally recombine into I, so the current through each resistor must be I /3 (the reverse argument from that of the first junction). If we follow one path from one corner to the other and add up the voltage drops on the resistors, the result must be the same as the voltage across the equivalent resistance of the cube. So, eq IR IR IR IR IR V = = + + = 6 5 3 6 3 and 6 5 R R eq = .
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.

This note was uploaded on 06/08/2010 for the course PHYS 213 taught by Professor Perelstein during the Spring '08 term at Cornell University (Engineering School).

Page1 / 6

HW5solutions-10 - Physics 2213 HW #5 Solutions Spring 2010...

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