# Version 001 CIRCUITS holland (1290) 1

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1 Version CRCUTS hollnd (9) This print-out should hve questions Multiple-choice questions my continue on the next column or pge find ll choices efore nswering AP M 99 MC points The power dissipted in wire crrying constnt electric current my e written s function of, the length l of the wire, the dimeter d of the wire, nd the resistivity ρ of the mteril in the wire n this expression, the power P dissipted is directly proportionl to which of the following? P d nd P ρ only P l only P l, P d nd P ρ P d only P l nd P ρ only correct According to Ohm s lw, we know the power dissipted is P R ρ l πd Therefore, the power is directly proportionl to P l nd P ρ only AP M 99 MC 67 points A vrile resistor is connected cross constnt voltge source Which of the following grphs represents the power P dissipted y the resistor s function of its resistnce R? correct The power dissipted in the resistor hs

2 Version CRCUTS hollnd (9) severl expressions P R R, wherethelsttworesimplyderivedfromthe first eqution together with the ppliction of the Ohm s lw Since the resistor is connected to constnt voltge source constnt P R constnt, R tells us tht the power( is inversely proportionl to the resistnce P ) R AP M 998 MC points A wire of resistnce R dissiptes power P when current psses through it The wire is replced y nother wire with resistnce R The power P dissipted y the new wire when the sme current pssed through it is P P correct P 6P AP B 99 MC (prt of ) points A ttery with n internl resistnce is connected to two resistors in series 6 Ω Ω A internl resistnce Wht is the emf of the ttery? 8 V V correct V V 6 V Ω Let : R 6 Ω, R Ω, r Ω, nd A R R P P P P 9 P P The power dissipted y resistor is given y P R SoifwechngeRtoR,thepower dissipted will e chnged correspondingly to P (R ) (R) P internl resistnce The totl resistnce of the circuit is R totl r+r +R Ω+6 Ω+ Ω Ω, so the emf of the ttery is R totl ( A)( Ω) V r

3 Version CRCUTS hollnd (9) (prt of ) points Wht is the potentil difference cross the terminls nd of the ttery? V 8 V correct V V V V V 6 V V V The potentil difference cross the terminls of the ttery is or V r V ( A)( Ω) 8 V, V V (R +R ) ( A)(6 Ω+ Ω) 8 V 6 (prt of ) points Wht power P internl is dissipted y the Ω internl resistnce of the ttery? P internl W P internl 8 W P internl 6 W correct P internl W P internl 6 W The power dissipted y the r Ω internl resistnce is P internl r ( A) ( Ω) 6 W AP M 99 MC (prt of ) points The switch hs een open for long period of time R R V mmeditely fter the switch is closed, the current supplied y the ttery is V R V (R +R ) R R V R correct V R +R Before the switch is closed, there is no chrge on the cpcitor, so the voltge is zero cross the cpcitor t this time Becuse it is not possile to chnge the chrge on the cpcitor like step function (or the current should e infinitely lrge), immeditely fter the switch is closed, the voltgecross the cpcitor (nd R ) is still zero Therefore, the voltge cross R is V ; ie, think of the cpcitor s eing short-circuit for this instnt of time So the current supplied y the ttery, whichisthesmesthecurrentgoingthrough R, is V R 8 (prt of ) points A long time fter the switch hs een closed, the current supplied y the ttery is C S

4 Version CRCUTS hollnd (9) V R V R V R +R correct V (R +R ) R R After long time, the cpcitor hs een chrged nd remined stle Tht mens the current going through R is the sme s the current going through R ; ie, think of the cpcitor s eing open-circuit for this time So we cn write down the eqution V R + R, which gives the current s V R +R AP M 99 MC 7 9 (prt of ) points The circuit hs een connected s shown in the figure for long time V Ω Ω 6 µf 7 Ω Ω Wht is the mgnitude of the electric potentil C cross the cpcitor? C V C 7 V C 9 V S C V C V 6 C 8 V 7 C V 8 C V 9 C 8 V C 9 V correct R t R C t t R R S Let : R Ω, R 7 Ω, R Ω, R Ω, nd C 6 µf After long time implies tht the cpcitor C is fully chrged nd therefore the cpcitor cts s n open circuit with no current flowing to it The equivlent circuit is t R R R R t R t R +R Ω+7 Ω Ω R R +R Ω+ Ω Ω t R t V Ω A

5 Across R Across R R V Ω A t R ( A)( Ω) V R ( A)( Ω) V Since nd re mesured fromthesme point, the potentil difference cross C must e C V V 9 V C 9 V (prt of ) points fthetteryisdisconnected,howlongdoesit tke for the cpcitor to dischrge to t e of its initil voltge? t t / 78 µs t t / µs t t / 76 µs t t / 7 µs Version CRCUTS hollnd (9) nd where C Req eq R l R +R Ω+ Ω 6 Ω, R r R +R 7 Ω+ Ω Ω ) ( R eq + R l R r ( 6 Ω + ) Ω Ω Therefore the time constnt τ is τ R eq C ( Ω)(6 µf) 8 µs The eqution for dischrge of the cpcitor is Q t Q e t/τ, or t e t/τ e t t / µs 6 t t / 66 µs 7 t t / 6 µs 8 t t / 8 µs correct 9 t t / µs t t / µs With the ttery removed, the circuit is l l l R R C R R r r r Tking the logrithm of oth sides, we hve t ( ) τ ln e t τ ( lne) (8 µs)( ) 8 µs AP M 998 MC (prt of ) points The following digrm shows prt of closed electricl circuit Ω 7 Ω 9 Ω

6 Version CRCUTS hollnd (9) 6 Find the electric resistnce R of the prt of the circuit shown etween point nd R Ω R Ω R 9 Ω R Ω R 8 Ω correct 6 R Ω 7 R Ω 8 R 6 Ω 9 R 8 Ω R Ω R R R Let : R Ω, R 7 Ω, nd R 9 Ω Since R nd R re in series, their equivlent resistnce R is R R +R Ω+7 Ω 7 Ω R 7 Ω R 9 Ω Since R nd R re connected prllel, their equivlent resistnce R is + R +R R R R R R R R R R +R (7 Ω)(9 Ω) 7 Ω+9 Ω 8 Ω (prt of ) points When there is stedy current in the circuit, the mount of chrge pssing point per unit time is greter t point thn t point greterinthe9ωresistorthninthe7ω resistor correct the sme everywhere in the circuit greter in the Ω resistor thn in the 7 Ω resistor greter in the 7 Ω resistor thn in the 9 Ω resistor The mount of chrge pssing point per unit of time (the sme s the current t point) is not the sme everywhere, ut it is the sme t point s t point ; ie, it is the sme in the Ω resistor s in the 7 Ω resistor t is greter in the 9 Ω resistor thn in the Ω or 7 Ω resistor From Ohm s Lw R, we hve R 7 Ω R 9 Ω > When there is stedy current in the circuit,themount ofchrgepssing pointper unit time is greter in the smller 9 Ω resistor thn in the lrger 7 Ω resistor AP M 998 MC 6 points

7 Version CRCUTS hollnd (9) 7 A resistor R nd cpcitor C re connected in series to ttery of terminl voltge V Which of the following equtions relting the current in the circuit nd the chrge Q on the cpcitor descries this circuit? Q C R V +QC R V C dq dt R V Q R correct C V Q C R Kirchhoff s lw tells us the voltge drop cross ny closed circuit loop is zero We know the voltge drop cross the resistor is R nd tht cross the cpcitor is Q, so the C eqution descriing the circuit is V Q C R AP M 998 MC 7 points Which of the following comintions of resistors would dissipte W when connected to 6 V power supply? 6 Ω Ω Ω 6 Ω Ω 8 Ω 8 Ω 8 Ω 6 Ω Ω 8 Ω correct 6 Power is Ω Ω Let : P W nd 6 V P R eq, so the comintion of resistors dissipting W when connected to 6 V power supply will hve the equivlent resistnce of Consider R eq P (6 V) W R R 8 Ω R eq R +R 6 Ω+ Ω 6 Ω Consider R R + R eq R R so R eq Ω + Ω Ω Consider R R R Ω Ω 6 Ω R eq R +R +R 6 Ω+ Ω+8 Ω 8 Ω

8 Consider R R Version CRCUTS hollnd (9) 8 Ampère s lw Frdy s lw so R + + R eq R R R R eq 6 Ω + Ω + 6 Ω 8 Ω Consider R R R + R eq R R +R R eq 8 Ω 6 Ω + Ω+ Ω Consider R 6 R 6 R 6 R eq R R 6 R 6 6 Ω+ Ω + Ω Ω AP B 99 MC points Kirchhoff s loop rule for circuit nlysis is n expression of which of the following? Ohm s lw Conservtion of energy correct Conservtion of chrge Kirchhoff s loop rule V V +V +V + follows from the conservtion of energy AP B 99 MC 6 6 (prt of ) points Consider the circuit µf µf c V µf µf Wht is the equivlent cpcitnce for this network? C equivlent 7 µf C equivlent 7 µf correct C equivlent µf C equivlent 7 µf C equivlent µf Let : C µf, C µf, C µf, C µf, nd B V

9 Version CRCUTS hollnd (9) 9 C 8 points B C c C C Consider resistors R nd R connected in series R R C nd C re connected in prllel, so C C +C 6 µf C nd C re connected in series, so + C +C C C C C C C C C C +C (6 µf)( µf) 6 µf+ µf µf C nd C re connected in prllel, so C C +C 7 µf 7 (prt of ) points Wht is the chrge stored in the -µf lowerright cpcitor? Q, µc Q 6 µc Q,8 µc Q µc correct Q 7 µc Let : C µf nd B V The chrge stored in cpcitor is given y Q CV, so Q C V ( µf)( V) µc AP B 99 MC nd in prllel R R to source of emf tht hs no internl resistnce How does the power dissipted y the resistors in these two cses compre? t is greter for the series connection tisdifferent for echconnection, ut one must know the vlues of to know which is greter tisdifferent for echconnection, ut one must know the vlues of R nd R to know which is greter t is the sme for oth connections t is greter for the prllel connection correct The power dissipted y the resistors is P R eq The equivlent resistnce for series connection is R s R +R The equivlent resistnce for prllel connection is R p R R R +R

10 Version CRCUTS hollnd (9) RegrdlessofthevluesofR ndr,r p < R s, so morepower isdissipted in theprllel connection AP M 99 MC 9 9 (prt of ) points Consider the system of equivlent cpcitors µf µf C C C C C C 6 µf µf µf µf We hve series comintions, ech with equivlent cpcitnce C C + C C, of B the cpcitors connected in prllel, so Find the equivlent cpcitnce C of the network of cpcitors C µf C µf C µf C µf C µf 6 C 6 µf 7 C µf correct 8 C µf C C + C + C C µf µf (prt of ) points Wht potentil difference must e pplied etween points nd so tht the chrge on ech plte of ech cpcitor will hve mgnitude of 6 µc? V V V V V V V V V V 6 V 6 V correct 7 V 9 V 8 V 8 V Let : Q 6 µc nd B 6 V Let : C C C C C C 6 C µf Regrd the system s n equivlent cpcitor with the cpcitnce C The chrge on

11 Version CRCUTS hollnd (9) series set of cpcitors is the sme: Q Q Q 6 6 µc, nd the chrge on prllelsetofcpcitorsisthesum ofthechrges on ech rnch: Q totl (6 µc) 8 µc, so C C C C c V Q totl C 8 µc µf 6 V AP M 99 MC (prt of ) points Consider the following system of equivlent cpcitors µf µf B d C C 6 We hve prllel comintions, ech with equivlent cpcitnce C C +C C, of the cpcitors connected in series, so c µf µf Find C d C + C + C C C µf ( µf) µf B µf the equivlent cpcitnce of the circuit C µf C µf correct C µf C µf C µf 6 C µf 7 C µf (prt of ) points Wht potentil difference must e pplied cross the cpcitor network so tht the chrge on ech plte of ech cpcitor will hve mgnitude of 6 µc? V d 6V correct V d V V d V V d V V d 8V 6 V d V 7 V d V 8 V d V Let : C C C C C C 6 C µf Let : Q 6 µc nd B 6 V

12 Version CRCUTS hollnd (9) Regrd the system s n equivlent cpcitor with the cpcitnce C d µf The chrge on ech prllel set of cpcitors is the sum of the chrges on ech of the cpcitors: Q Q Q 6 µc, nd the chrge on series set of cpcitors is the sme: Q totl µc, so R R R V d Q totl C d µc µf 6 V AP M 998 MC 6 (prt of ) points The following digrm shows prt of closed electricl circuit Ω 9 Ω 99 Ω FindtheelectricresistnceR oftheprt of the circuit shown etween point nd R Ω R Ω R 7 Ω R Ω R 9 Ω 6 R Ω 7 R 7 Ω 8 R 9 Ω 9 R 9 Ω R 7 Ω correct Let : R 9 Ω, R Ω, nd R 99 Ω Since R nd R re connected prllel, their equivlent resistnce R is + R +R R R R R R R R R R +R R 9 Ω ( Ω)(99 Ω) Ω+99 Ω R 8 Ω 8 Ω Since R nd R re in series, their equivlent resistnce R is R R +R 9 Ω+8 Ω 7 Ω R 7 Ω (prt of ) points When there is stedy current in the circuit, the mount of chrge pssing point per unit time is greter t point thn t point greter in the Ω resistor thn in the 9 Ω resistor greter in the 99 Ω resistor thn in the 9 Ω resistor

13 Version CRCUTS hollnd (9) greter in the Ω resistor thn in the 99 Ω resistor correct the sme everywhere in the circuit The mount of chrge pssing point per unit of time (the sme s the current t point) is not the sme everywhere, ut it is the sme t point s t point, nd it is the sme in the 9 Ω resistor s in the comintion of the prllel R + R 8 Ω resistors Therefore the current is less in the Ω resistor or 99 Ω resistor thn in the 9 Ω resistor, nd it is greter in the Ω resistor thn in the 9 Ω or 99 Ω resistor From Ohm s Lw R, we hve R Ω R 99 Ω > When there is stedy current in the circuit, the mount of chrge pssing point per unit time is greter in the smller Ω resistor thn in the lrger 99 Ω resistor

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