Drill 5
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University of Central Oklahoma *
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2114
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Electrical Engineering
Date
Apr 3, 2024
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© Add one more victory to Central’s history! © Name: /\ 0 \/\\\'M éC H?M Drill #5, Spring 2023. MWF 9-9:50 -/ / Page 1 of 4 AA cells have an EMF of 1.50 V. When a current of 0.250 A is drawn from the cell, its terminal voltage is measured to be 1.30 V. 1. [5 pts] Sketch cell, including the EMEF, the current, and the internal resistance. r-“u’ 2. What is the internal resistance of the AA cell? Identify Variables and Constants. Write “?” if the value is initially unknown. Required. Symbol Description Value Units VU/EMF = Electromotive force of AA cell = 1.50 v K = Terminal voltage of AA cell Y= 1.30 v T = Current drawn from AA cell = 0.250 A R = Internal resistance of AA cell = ? Q Accepted Equations [3 pts] Show your work here. Keep the units at each step! [5 pts] U= TR /U; - @/gmgf - Kin: T Q\;‘* - Uerr —;JT/I ALy - 0.y Symbol Value Units Final answer. [2 pts] K. = 0.¢ L+ Q
© Add one more victory to Central's history! © Name: Drill #5, Spring 2023. MWEF 9-9:50 Page 2 of 4 Paul Peter Jingleheimer-Schmidt is installing a lightning rod on the side of his house. The lightning rod is 7.29 m long and has a circular cross-section with a radius of 2.00x102 m. The lightning rod is made of stainless steel, which has a resistivity of 69.0x10® Q-m. 3. [5 pts] Sketch the situation described above, showing the radius and length of the rod. PN 4. What is the resistance of the lightning rod? Identify Variables and Constants. Write “?” if the value is initially unknown. Required. Symbol Description Value Units L = Length of lightning rod = 7.29 m = Radius of lightning rod = 2.00x1072 m Fool= Resistivity of stainless steel = 6.90x107 Q'm Q = Resistance of lightning rod = ? Q Accepted Equations [3 pts] Show your work here. Keep the units at each step! [5 pts] - PL AL (Wox 0T em) (T 24 Y s R= & Q- W,%.ooxm' o\ T A= pv = 4.00xIp 3.0 Final 2 pts] Symbol Value Units inal answer. [2 pts , —7 R =1 4000 Q
®© Add one more victory to Central's history! @ Name: Drill #5, Spring 2023. MWEF 9-9:50 Page 3 of 4 A 500-22 resistor is in series with another resistor. The voltage across the combination of resistors is 1.00x10? V. The total power dissipated is 8.00 W. 5. [5 pts] Sketch the two resistors in series and the voltage across the combination. AN 0 VK Y 2 i /\ AAA A 'f\v’\/’\“ —— 7= "b[ 6. What is the resistance of the other resistor? - Identify Variables and Constants. Write “?” if the value is initially unknown. Required. Symbol Description Value Units R, = Resistance of resistor #1 = 5.00x10? Q “\// = Voltage across combination of resistors = 1.00x10? v @ = Total power dissipated = 8.00 W R. |= Resistance of resistor #2 = ? Q Accepted Equations [3 pts] Show your work here. Keep the units at each step! [5 pts] QSU‘\LS N ip\r\ R=vp = QO0XI0" Y /iy = 1230 P-7"JQ ‘Q‘KU.\LS N V\l ) @\’L Q‘l = R - g\ - WS0 9 -S0p S = 700 O Symbol Value Units Final answer. [2 pts] Re =] 2
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® Add one more victory to Central’s history! © Name: Drilt#5, Spring 2023. MWF 9-9:50 Page 4 of 4 A 500-L2 resistor is in parallel with another resistor. The total current through the combination of resistors is 2.00x10* A. The total power dissipated is 8.00 W. 7. [5 pts] Sketch the two resistors in parallel and the total current through the combination. 8. What is the resistance of the other resistor? Identify Variables and Constants. Write “?” if the value is initially k41 Lt v YNV = unknown. Required. Symbol Description Value Units Q\ = Resistance of resistor #1 = 5.00x10? Q Ty | = Current through combination of resistors = 2.00x10™ A 4 = Total power dissipated = 8.00 W R, |= Resistance of resistor #2 = ? Q Accepted Equations [3 pts] Show your work here. Keep the units at each step! [5 pts] @Qum\u - E‘E’T \@::\1@ T~ ’h = LW “/(1.00x\b“ A = 2000 | = t 2{){7)9\ /:S:m__(; n ) el vl Ty ?\1 - (O‘OOBS—’JL 3333% QL Symbol Value Units Final answer. [2 pts] B =] 233.>
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need part a, b and c asap i willl give you upvote
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ore
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so then how is Rth 0.8695 kOhms in the original explanation?
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