DRECT CURRENT CUTS ELECTRC POWER Consider the circuit shown in the Figure where bttery is connected to resistor R. A positive chrge dq will gin potentil energy s it moves from point to point b through the bttery. The chrge loses the sme mount of potentil energy s it moves from point c to point d through the resistor. This mount is given by d V R c b du = dqε = dqv Now du dq P = = V = V dt dt Using the formul V=R P = V = V R = R Exmple 7.7 You re given n electric heter mde of nichrome wire of resistnce 8 Ω. Find the current crried by the wire nd the power of the heter if it is connected to 0 V source. Solution = R V 0 = = 5A 8.0 The power cn be found using P ( 5) (8.0).8 KW = = R =
ELECTROMOTVE FOE Consider the circuit shown in the figure. Current will be estblished through the resistor R if potentil r b difference is mintined cross its ends. f the end b is t higher potentil thn the end, then chrge will move through the resistor from b to. For the current R to circulte round closed circuit, the chrge must jump from to b. This mens tht we need device tht cpble of pumping chrge from lower potentil to higher potentil. The function of such device is clled electromotive force, bbrevited emf, nd denoted by the symbol ε. The bttery nd the genertor re common emf devices. A source of emf cn be considered s chrge pump tht pumps chrges in direction opposite to the electrosttic force inside the source. t is exctly like wter pump tht pushes wter from lower to higher level opposite to the grvittionl force. The resistnce r is clled the internl resistnce of the bttery, nd R is clled the lod resistor. We shll ssume tht the connecting wire hve no resistnce. Any positive chrge moving from to b will gin potentil ε s it psses from the negtive terminl to the positive terminl of the bttery. However, it will lose potentil r s it psses through the internl resistor, where is the current in the circuit. Thus, the terminl voltge of the bttery, V = V b V, is given by V =ε r 8. From this Eqution it is cler tht the emf is equl to the terminl voltge of bttery in n open circuit, tht is when the current is zero. As the connecting wires hve no resistnce we conclude tht the voltge V must lso equl the potentil cross the lod resistor R, tht is V = R ε
From the bove two Equtions we get = ε R + r 8. Exmple 8. A bttery hs n emf 0f.0 V nd n internl resistnce of 0.05 Ω. ts terminls re connected to lod resistnce of 3.0 Ω. ) Find nd the terminl voltge of the bttery. b) Clculte the power delivered to R, r, nd by the bttery. Solution: ) ε.0 = = = 3.93A R + r 3.0 + 0.05 V = ε r =.0 (3.93 0.05) =.8V b) = R= ( 3.93) ( 3.0) = 46.3W P R P r = r = ( 3.93) ( 0.05) = 0.77W P ε = ε = 3.93.0= 47.W Note tht P = P R + P ε r
RESSTORS N SERES AND N PARALLEL When two resistors re connected together s shown we sid tht they re connected in series. As it is cler from the figure, ny chrge tht flows through R must equl the chrge tht flows through R, tht is the current is the sme in ech resistor. Since the potentil difference between nd b equls the sum of the potentil drop cross ech resistor we hve R ε R b = R + R = ( R ) 8.3 V eq + R where V eq is the potentil drop cross the equivlent resistor. Therefore we conclude tht R eq = R + R 8.4 The equivlent resistor of more thn two resistors connected in series is then R eq R + R + +L 8.5 = R3 R Now consider the two resistors connected s shown in the Figure. The potentil drops cross R nd R re equl nd must equl to the potentil drop cross ny equivlent resistor connected between nd b, tht is R b V eq = V = V 8.6 ε f nd re the currents pssing through R nd R, respectively, then the net current of the circuit is
= + 8.7 Using Ohm s lw nd Eqution 8.6, we get R = + eq R R 8.8 n generl if more tht two resistors re connected in prllel, then we hve = + + +L 8.9 R eq R R R 3 Exmple 8.4 Four resistors re connected s shown in. () Wht is the equivlent resistnce between points & c.? (b) Wht is the current in ech resistor if potentil difference of 4 V is mintined between & c.? 8.0 Ω 4.0 Ω 3.0 A 3.0 A b.0 A 6.0 V 6.0 Ω 3.0 Ω 6.0 V.0 A c Ω.0 Ω 3.0 A b 3.0 A c 36 V 6.0 V 4 Ω 3.0 A 4 V c
Solution: () The circuit cn be reduced, step by step, to single equivlent resistnce s shown in the Figure. The 8.0-Ω nd the 4.0-Ω re connected in series, nd so they cn be replced by n equivlent resistor of Ω. The 6.0-Ω nd the 3.0-Ω re connected in prllel, nd so they cn be replced by n equivlent resistor of.0 Ω. The equivlents re connected in series. The equivlent resistnce of the circuit is then R eq = +.0 = 4Ω (b) Since the -Ω nd the.0-ω re connected in series, they hve the sme current eq, which must equl to the current of the 4-Ω resistor. Using Ohm s lw we get R eq eq = Req 48 = = 3.0A 4 Now the potentil difference cross the.0-ω is V bc = R = 3.0(.0) = 6.0 V eq This potentil difference is the sme cross the 6.0-Ω nd the 3.0-Ω resistors due the prllel connection between them. So, we cn find the current pssing through the 3.0-Ω resistor s 6.0 = = 3.0.0A nd the current through the 6.0-Ω resistor s 6.0 = =.0 A 6.0 The current pssing through the 4.0-Ω nd the 8.0-Ω is the sme s tht pssing through the -Ω (3.0 A) due the series connection between them.
Exmple 8.5 shown? Wht is equivlent resistnce between nd b in the Figure c Ω 5 Ω Ω d Ω Ω b 5 Ω Ω Ω Ω c,d Ω b 0.5 Ω 0.5 Ω c,d b 0.5 Ω b Solution There re no series or prllel connections in the system given. Now consider current entering the junction. Becuse of the symmetry in the circuit, the current in brnches c & d must be equl, nd hence the points c & d hve the sme potentil (V cd =0), tht is the circuit cn be reduced s in the figure.
Exmple 8.6 Three resistors re connected in prllel s shown. A potentil difference of 8 V is pplied cross points nd b. ) Find the current in ech resistor. b) Clculte the power delivered to ech resistor nd the totl power delivered to the combintion. 8 V 3Ω 6 Ω 3 9 Ω Solution: ) Since the resistors re connected in prllel, the potentil difference cross ech one is the sme nd equl to 8 V. Now V 8 = = = 6A R 3 V 8 = = = 3A R 6 V3 8 3 = = = A R 3 9 b) For the power delivered to ech resistor we pply P = R = ( 6) ( 3) = 08 W P = R = ( 3) ( 6) = 54 W P 3 = 3 R3 = ( ) ( 9) = 36 W P R = ( ) (.64) = 98 W eq = eq eq Note tht P eq = P + P + P3 b
8.3 KHHOFF S RULES - The sum of the currents entering ny junction must equl the sum of the currents leving tht junction. (A junction is ny point in circuit where current cn split) - The lgebric sum of the potentil differences cross ll the elements round ny loop must be zero. The first rule is n ppliction of the conservtion of chrge principle, while the second rule is n ppliction of the conservtion of energy principle. To pply the second rule we should know the following two remrks: - The chnge in potentil through ny resistor is negtive for move in the direction of the current nd positive for move opposite to the direction of the current. This is becuse the current through resistor moves from the end of higher potentil to tht of lower potentil. - The chnge in potentil through n idel bttery is positive for move from the negtive to the positive terminl of the bttery nd negtive for move in the opposite direction. Strtegy for solving problems using Kirchhoff s rules: - Drw circuit digrm nd lbel ll quntities, known nd unknown. - Assign direction for the current in ech prt of the circuit. Do not bother if your guess of current direction is incorrect; the result will hve negtive vlue. 3- Apply the first Kirchhoff s rule to ny junction in the circuit. n generl this rule is used one time fewer thn the number of junctions in the circuit. 4- Choose ny closed loop in the network, nd designte direction (clockwise or counterclockwise) to trverse the loop. 5- Strting from one point in the loop, go round the loop in the designted direction. Sum the potentil differences cross ll the elements of the chosen loop to zero. n doing so you should note the two remrks discussed bove, tht is, the potentil difference cross n emf is +ε if it is trversed from the negtive to the positive terminl nd -ε if trversed in the opposite direction. The potentil
difference cross ny resistor is -R if this resistor is trversed in the direction of the ssumed current nd +R if trversed in the opposite direction. 6- Choose nother loop nd repet the fifth step to get different eqution relting the unknown quntities. Continue until you hve s mny equtions s unknowns. 7- Solve these equtions simultneously for the unknowns. Exmple 8.8 n the circuit shown, find the current in the circuit nd the power delivered to ech resistor nd the power delivered by the -V bttery. ε =6V R = 0 Ω R = 8 Ω Solution The directions of the currents re ssigned rbitrry s shown in the Figure. As it cler from the circuit there is one loop with no junctions. Now, we pply Kirchhoff s second rule to the loop nd trverse the loop in the clockwise direction, obtining ε ε R R = 0 6 8 0 = 0 6 6 = = = 0.33A 8 + 0 8 The minus sign indictes tht the direction of is opposite the ssumed direction. To find the power delivered to ech resistor, we use = R = = R = P P 0.87 W. W ε =V
And for the power delivered by the bttery we hve P = ε = 0.33 = 4W Note tht P + P = 0.87 +. = W Tht is hlf of the power supplied by the V-bttery is delivered to the resistors nd the other hlf is delivered to the 6V-bttery Exmple.4 ) Find the current in ech resistor in the figure shown. b) Clculte the potentil difference Vb V. 4V Solution ) f we pply Kirchhoff s first rule to the junction b we get 4Ω 6 Ω b = + 3 () Now pplying Kirchhoff s second rule to the upper loop trversing it clockwise we get 0V Ω 3 4 + 6 0 4 = 0 () For the bottom loop trversing it in the clockwise direction gives 0 6 3 = 0 (3) Substituting for 3 from Eq. () into Eq.(3) 0 6 ( + ) = 0 0 8 = 0 (4) Dividing Eq.() by + 3 = 0 (5) Subtrcting Eqs. (4) & (5) = 0 or =.0 A
From Eq. (4) we get or = 3.0A nd from Eq. () we get or 3 =.0A The minus sign indictes tht & 3 should be reversed b) Strting t point, we follow pth towrd point b, dding potentil differences cross ll the elements we encounter. f we follow the pth through the middle bttery we obtin V V b = 0 6 = 0 6 = V The minus sign here mens tht V > V b. Try to follow nother pths from to b to verify tht they lso give the sme result. The CUTS Chrging Process The figure shows cpcitor, initilly unchrged, connected in series with ε C resistor. f the switch S is thrown t point t t = 0, the cpcitor will begin to chrge, creting current in the circuit. Let be the current in the circuit t some instnt during the chrging process, nd q be the chrge on the cpcitor t the sme instnt. Applying Kirchhoff s second rule to the circuit, we obtin q ε R = 0 8.0 C S R Substituting for with dq = εc q dt = dq dt, in Eqution 8.0, nd rerrnging we obtin 8.
Noting tht the chrge on the cpcitor is initilly zero, i.e., q = 0 t t = 0, we cn integrte both sides of Eqution 8. s q dq = C q 0 ε t 0 dt εc q ln = εc t t q = ε C e 8. where e is the bse of the nturl logrithm. To find the current s function of time, we differentite Eqution 8. with respect to time to get t = ε e 8.3 R The quntity is clled the time constnt, τ, which defined s the time required for the current to decrese to e of its initil vlue. Equtions 8. nd 8.3, which re plotted in the following Figure, tell the following: - At t = 0, the chrge q is zero, s required, nd the initil current o is
q Q m o t t () (b) Figure.8 () The chrge versus time in chrging process for circuit. (b) The current versus time in chrging process for the sme circuit. o ε = 8.4 R tht is, the cpcitor cts s if it were wire with negligible resistnce (short circuit). - As t (fter long time), the chrge hs its mximum equilibrium vlue, Q m Q m = ε C 8.5 nd the current is zero, tht is the cpcitor cts s it were n open switch (open circuit). Dischrging Process Suppose tht the cpcitor is now fully chrged such tht its potentil difference is equl to the emf ε. f the switch is thrown to point t new time t = 0, the cpcitor begin to dischrge through the resistor. Let be the current in the circuit t some instnt during this process, nd q be the chrge on the cpcitor t the sme instnt. Applying Kirchhoff s rule to the loop, we get
q R = 0 C Substituting for with obtin 8.6 = dq dt (explin the negtive sign), nd rerrnge we dq = dt 8.7 q Using the initil condition, q = Q m t t = 0 we cn integrte the lst eqution to obtin q Q ln mx q Qm dq t = dt q 0 = t q t = Q m e 8.8 The current is the rte of decrese of the chrge on the cpcitor, tht is dq dt t o e = = 8.9 where Q o = 8.0 Exmple 8. A 8 0 5 -kω resistor nd 5-µF cpcitor re connected, in series, with -V bttery s shown. The cpcitor is initilly unchrged, nd the switch S is closed t t=0.
) Find the time constnt of the circuit, nd the mximum chrge on the cpcitor. b) Wht is the time required for the current to drop to hlf its initil vlue? c) After being closed for long time, the switch is now opened t t=0, wht is the time required for the chrge nd for the energy to decrese to one-fourth their mximum vlue. V S R 5 µf Solution ) The time constnt is τ = = 5 6 ( 8.0 0 )( 5.0 0 ) = 4.0s The mximum chrge is, Q m 6 ( )( 5.0 0 ) 60µ C = ε C = = And the mximum current is ε o = = 5µ A R 5 8 0 = b) Now we hve = o e t To find the time required for the current to drop to hlf its vlue, we substitute = into this eqution: o o = o e t Tking the logrithm of both sides, we hve
or t ln = t = ( ln ) =.8s c) n the dischrging process, the chrge vries with time ccording to q = Q m e t Substituting for q = Q, nd tking the logrithm of both sides we get 4 m or ln 4 = t = t ln 4 = 5.5s For the energy we hve q Q t t U = = e = U m e C C Substituting for U = 4 U m, nd tking the logrithm of both sides we get or ln 4 = t = t ln = 4.8s
Exmple (Extr) n the circuit shown, the cpcitor is initilly empty nd the switch S is closed t t=0. ) Find the current in ech brnch of the circuit t t=0. b) Clculte the mximum chrge on the cpcitor. 3V 6 Ω S µf 3 4Ω Solution ) At t=0, the cpcitor is treted s if it were wire with negligible resistnce. This mens tht the cpcitor mkes short circuit cross the 4-Ω resistor. Therefore, we hve = 0 nd 3 = 3 = = 6.0 5.3 A b) The mximum chrge is ttined fter long time ( t ). At this time the cpcitor is treted s if it were n open switch. So, 3 = 0 nd 3 = = = (6.0 + 4.0) 3.A. To clculte the chrge on the cpcitor, we first wnt to find the potentil difference cross it. Applying Kirchhoff s second rule to right loop we find tht the potentil difference V cross the cpcitor is V = R ( 3.)( 4.0) V = = Now the mximum chrge is ( 6 0 )(.8) =.5 0 4 C Q = CV = µ