# Kirchhoff s Rules. Kirchhoff s Laws. Kirchhoff s Rules. Kirchhoff s Laws. Practice. Understanding SPH4UW. Kirchhoff s Voltage Rule (KVR):

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1 SPH4UW Kirchhoff s ules Kirchhoff s oltge ule (K): Sum of voltge drops round loop is zero. Kirchhoff s Lws Kirchhoff s Current ule (KC): Current going in equls current coming out. Kirchhoff s ules etween nd =- = () Lel ll currents Choose ny direction () Choose loop nd direction Must strt on wire, not element. Kirchhoff s Lws E E E 4 4 =+E =-E () Write down voltge drops -tteries increse or decrese ccording to which end you encounter first. -esistors drop if going with current. -esistors increse if gong ginst current. 5 e - - -e =0 For inner loop Lel Choose Write K currents loop or Find : ε - - ε - = = 0 = + mps Prctice e = 50 =5 W Wht if only went from to?, Find - - = e - = 50-5 = 40 olts e = 50 - = + + e = = +40 olts Therefore is 40 higher thn =5 W =5 W =5 W e = 0 e = 0 esistors nd re: Understnding ) in prllel ) in series ) neither Definition of prllel: Two elements re in prllel if (nd only if) you cn mke loop tht contins only those two elements. E = 5 =0 W =0 W + - E = 0 Definition of series: Two elements re in series if (nd only if) every loop t Contins lso contins

2 Prctice slide 7 Understnding slide 7 Clculte the current through resistor. ) = 0.5 ) =.0 ) =.5 e 0 e =0 W E = 5 =0 W <-- Strt E = 0 Understnding: oltge Lw How would chnge if the switch ws closed? ) ncrese ) No chnge ) Decrese 0 0W Clculte the current through resistor. ) = 0.5 ) =.0 ) =.5 E E W 0 5 0W 0.5 E = 5 =0 W =0 W E = 0 Strting t Str nd move clockwise round loop Kirchhoff s Junction ule Current Entering = Current Leving = + slide 7 () Lel ll currents Choose ny direction Kirchhoff s Lws Understnding ) = 0.5 = + ) =.0 = ) =.5 =.5 =.0 =0 W E = 5 =0 W = E = 0 () Choose loop nd direction Your choice! () Write down voltge drops Follow ny loops (4) Write down node eqution in = out E E E Ohm s Lw Series esistor Prllel Wiring oltge Ech resistor on the sme wire. Different for ech resistor. totl = + Ech resistor on different wire. Sme for ech resistor. totl = = Current Sme for ech resistor totl = = Different for ech resistor totl = + esistnce ncreses eq = + Decreses / eq = / + /

3 Understnding / Which configurtion hs the smllest resistnce? Which configurtion hs the lrgest resistnce? Prllel + Series Tests esistors nd re in series if nd only if every loop tht contins lso contins esistors nd re in prllel if nd only if you cn mke loop tht hs ONLY nd loop tht ONLY contins Understnding Determine the voltge nd current in ech resistor First we notice the voltge drop =5W through these =5W two resistor EQ groupings totl 0. =7W e 0=0 Let s comine to find EQ 4=5W 5W 5W 7W EQ 6.04W Let s now dd 4 to this EQ tot EQ 6.04W 5W 4.04W e 0=0 4 : 4 =5W =7W 0.9 =5W Understnding 4=5W 4 tot W 4.5 Since voltge drop hs to totl 0, EQ =0-4.5=5.5 e 0=0 T=.04W W,,, experience the sme voltge EQ : : : 5.5 5W EQ 5.5 5W Let s determine the current, T This current will flow through ech of the resistor groupings EQ 5.5 7W The -- Chrt Determine the oltge, Current, nd esistnce 7Ω The -- Chrt Determine the oltge, Current, nd esistnce T 5W 4.7W 9.7W 7Ω 7W 0W eq 4.7W eq Step : Fill out the tle with known resistors nd the Totl oltge for circuit Totl Step : Using resistor lws, determine totl resistnce of circuit Totl 9.7

4 The -- Chrt Determine the oltge, Current, nd esistnce 9.7W 7Ω.. 5W 6.6 The -- Chrt Determine the oltge, Current, nd esistnce 7Ω Step 4: Step : Since initil current mount will Using lso pss Ohm s through lw, determine resistor, the we Current cn determine of circuit. its voltge drop Totl. 9.7 Step 5: Since nd hve the sme oltge drop, we then must hve -6.6= Totl. 9.7 The -- Chrt Determine the oltge, Current, nd esistnce 7Ω Step 6: Use ohm s lw to find currents Totl. 9.7 f 60 wtt light ul opertes t voltge of 0, wht is the resistnce of the ul? ) Ω ) 0Ω c) d) 70Ω e) 700Ω P 0 40 P 60W W Three resistors re connected to 0- ttery s shown elow. Wht is the current through the.0 Ω resistor? ) 0.5 ) 0.50 c).0 d).0 e) 4.0 ε=0 Since ll resistors re in series, the mount of current tht psses through ny one of them is the sme. So we need to simply the circuit to determine tht current. 4.0Ω s... n 4W 4W W 0W 4.0Ω.0Ω 0 0W s... n Determine the equivlent resistnce etween points nd? ) 0.67Ω W ) 0.5 Ω W 4W P c) 0. Ω d).5 Ω W W 6W e) Ω Ω W 6W W 4Ω Ω Ω 4

5 Wht is the voltge drop cross the ohm resistor in the portion of the circuit shown? ) 4 ) 6 c) 48 d) 7 e) 44 8Ω 4Ω 8Ω Ω W 8W 4W 8W P W W 4W Since the top rnch hs 4Ω of resistnce nd the ottom rnch hs Ω of resistnce, therefore times s much current will flow through the 4Ω resistor thn the Ω resistor. This reks down the into 9 up nd down. So form = we hve =()(Ω)=6 Ω We could lso hve produced T nd found Ω vlue, therefore voltge drop of (Ω)()=6 on oth the upper nd lower rnch 9 00Ω,, nd 50Ω resistor re connected to 9- ttery s in the circuit shown elow. Which of the three resistors dissiptes the most power? We cn lso nswer this question fster P ) The 00Ω resistor y noticing tht the voltge drops cross ) The resistor the 00Ω nd then cross the prllel c) The 50Ω resistor resistors. P Since Pthe 00Ω hs P higher d) oth the nd the 50Ω vlue thn the prllel comintion, it will e) ll dissipte the sme power hve 5.4 lrger voltge.6drop thn the.6 comintion so vi P=, it dissiptes the 00Ω 00W 0W 50W most power. 0.9W 0.08W 0.086W 50Ω Therefore the 00Ω resistor dissiptes the most mount of power. ) t wht rte does the ttery deliver energy to the circuit? ε=0 0Ω 00Ω ) t wht rte does the ttery deliver energy to the circuit? ) Determine the current through the 0 Ω resistor. c) i) Determine the potentil difference etween points nd ii) t which of these two points is the potentil higher? d) Determine the energy dissipted y the 00 Ω resistor in 0 s ecll: te is Power P We need to write this s simple circuit with T so tht we cn determine the current, y using =. T 0W 0W 0W 40W 0 00W 60W 0 60W T ε=0 P 0 40W 0Ω 00Ω ecll: from ) we hve current The rtio of the resistnce of left side to right side is 40:0 or : ) Determine the current through the 0 Ω resistor. ε=0 0Ω 00Ω Therefore ¼ of goes through the right nd ¾ of goes through the left. Thus (/4)()=0.5 psses through the 0Ω resistor. We cn determine the voltge drop. 0- ()()- ()()- ()() =60 Then we use = to find ech individul current. c) i) Determine the potentil difference etween points nd ii) t which of these two points is the potentil higher? i) 0 00 ε=0 0.50W 0.500W 60 0Ω 00Ω ii) Point is t higher potentil. Since current flows from high potentil to low potentil. O 5

6 d) Determine the energy dissipted y the 00 Ω resistor in 0 s ecll: Energy equls Power multiplied y Time E Pt P ε=0 E t W0s 50J 0Ω 00Ω Let s follow conventionl current pth through T e W 0.95W Understnding e 0=0 =5W =7W =5W You cn pick ny pth through the circuit nd the totl voltge increses nd decrese will lnce 4=5W You cn reverse the direction of the current nd thus the signs, (tteries increse the voltge, resistors drop the voltge) nd otin the sme results. Let us clculte the Current nd the Power (used/generted) y the elements of the following circuit. Wht hppens to the Power delivery nd consumption if nother identicl ul is plce in prllel or in series with the first? P.5W 5 5.5W.5W 0W 5.5W e 5 P 5 ) 0W Let us clculte the Current nd the Power (used/generted) y the elements of the following circuit when uls re in prllel. P P.5W 5.5W 5 5.5W.5W 5 0W.5W.5W 0W 5.5W e 5 P 5 ) 0W ecuse the uls (resistors) re in prllel, we use the prllel lw to determine totl resistnce of the circuit W 4.5W e 5 P 45 ) 0W Let us clculte the Current nd the Power (used/generted) y the elements of the following circuit when uls re in series. P.5W P.5W W.5.5W 0W.5W.5W ε= 9 00Ω 5.5W e 5 P 5 ) 0W ecuse the uls (resistors) re in series, we use the series lw to determine totl resistnce of the circuit. 5 5 W.5W.5W e 5 P 5) 5W ) Simplify the ove circuit so tht it consists of one equivlent resistor nd the ttery. ) Wht is the totl current through this circuit? c) Find the voltge cross ech resistor. d) Find the current through ech resistor. e) The resistor is now removed from the circuit. Stte whether the current through the 00Ω resistor would increse, decrese, or remin the sme. 6

7 00Ω 00Ω ε= 9 EQ ε= 9 EQ =4.Ω ) Simplify the ove circuit so tht it consists of one equivlent resistor nd the ttery. 400W 500W 000 W 9.W 00W 00W 0W 000 EQ W 0W 9 4.W EQ ) Wht is the totl current through this circuit? 9 4.W Ω 00Ω c) Find the voltge cross ech resistor. d) Find the current through ech resistor. Let s use chrt ε= 9 e) The resistor is now removed from the circuit. Stte whether the current through the 00Ω resistor would increse, decrese, or remin the sme. ε= 9 00Ω Ω Totl Ω 00W 00W 0W W y removing resistor from prllel set, we ctully increse the resistnce of the totl circuit. Therefore y Ohm s lw if the voltge remins the sme nd the resistnce increses, the totl current must decrese Now through the 00Ω set, the totl resistnce remins the sme, yet the current decreses, therefore the voltge cross ech resistor decreses s well s the current. 7

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