Kirchoff s Laws and Their Use for Circuit Analysis Equations s i V=I i P=IV p i i Kirchoff s Laws Loop Law The total potential change around a closed circuit equals zero. Current Law for a Point For an point in a circuit the current into the point is equal to the current out of the point. A. For the following circuits draw the equivalent circuit and (you will need to work on a separate paper) a. calculate the total resistance of the circuit b. calculate the total current through the circuit c. calculate the current through each resistor d. calculate the voltage across each resistor e. calculate and the power dissipated by the entire circuit. V I P = 6 Ω V = 8 volts % i =-/3 amps @ 4 W = 3 Ω V = 4 volts & i =-/3 amps @ 7 W T = 9 Ω * VT = V it =-/3 amp + W Total esistance Add the resistors in series to get total resistance of the circuit: + = 9 ohms * (Use the symbols next to the numbers to show the order the numbers are filled in the table above. The asteric (*) is the first value calculated) Calculate the Total Current = 9 Ω, V = Volts V = I solve for I ; I = V/ V/9Ω = -/3 amp + Current Through Each esistor Because the resistors are in series they have the same current. This is the same as the total current -/3 amps @. i = i = I total Voltage Drop Across Since the current through the resistor and the value of the resistor are known the voltage drop can be calculated with Ohm s Law; V = I V = (-//3 amp)(6 Ω) = 8 volts % Voltage Drop Across
The same method used to find the voltage across could be used for however it is a good opportunity to illustrate the use of The Loop ule. The total potential change around a closed circuit equals zero. The battery increases the voltage by volts and the first resistor decreases it by 8 volts, leaving 4 volts to be dropped. V battery V V = 0 or V = V battery V = volts 8 volts = 4 volts & Power For Each esistor The power for each resistor is calculated with P = IV. The total power is the sum of all the powers. This calculation in not shown here.
B V I = 8 Ω V = 6V i = ¾ A A = 4 Ω V = 6V i = 6/4 A T =-/3 VT = 6V it = 9/4 A See what can be figured from the diagram before using any of Kirkoff s Laws. ** LOOP LAW The resistors have the same voltage drop across them and this is the same as the total voltag created by the generator, 6 Volts. ** CUENT LAW IT = i + i Total esistance (from previous homework) total = -/3 Ohms Total Current Total Current can be found with Ohm s Law: V = I so I = V/, plugging in gives Itotal = (Vtotal/total) = (6 V / -/3 Ω ) = 9/4 Amps Current in the Individual esistors Total Current can be found with Ohm s Law: V = I so I = V/, plugging in gives i = (V/) = (6 V / 8 Ω ) = 3/4 Amps It is often advantageous to leave the values as fractions. i = (V/) = (6 V / 4 Ω ) = 6/4 Amps Checking Answers with Current Law The currents may be checked using the Current Law. For Point A on the diagram I = 0: IT i i = 0 9/4 Amp - ¾ Amp 6/4 Amp = 0 C
V I = 0 Ω V = V i =. A = 80 Ω V = 3 V i =.07 A 3 = 0 Ω V3 = 6.6 V i3 =.08 A 4 = 50 Ω V4 = 7V i4 =.08 A T = 40 Ω VT = 4V IT =. A See what can be figured from the diagram before using any of Kirkoff s Laws. ** The current through the 0 ohm resistor is equal to the total current: Itotal = i ** 3 and 4 are in series so they have the same current: i3 = i4 Total esistance (from previous homework) = 40 ohms total = 40 ohms Total Current Total Current can be found with Ohm s Law: V = I so I = V/, plugging in gives Itotal = (Vtotal/total) = (4 V / 40 Ω ) =. amps V - Voltage across The voltage across can be found with the total current (Itotal = i) and Ohm s Law: V = I, plugging in gives V = (i / ) = (. amp x 0 ohm) = Volts V Voltage across The voltage across can be found with Kirkoff s Loop (or Voltage Law). The Loop Law: The total voltage gained is equal to the total voltage dropped around any loop Vbattery V V = 0 so V = 4V V = 3 Volts Find i
The current through can be found with Ohm s Law. V = I so I = V/ i = V / = (3 V / 80 ohm) =.07 amps Find Current in 3 and 4 Because these resistors are in series they share the same current: i3 = i4 Use Kirkoff s Current Law on Point A in the diagram. The Current Law: The current into any point in a circuit equals the current out of the point The following Equation can be written for point A: i i i3 = 0 so i3 = i i =. amp -.07 amp =.08 amp Find the Voltage Drop for Find Current in 3 and 4 Ohm s Law is used to find the voltage for each resistor: V = I V3 = i33 = (.08 A)(0 ohm) = 6.6 Volts V4 = i44 = (.08 A)(50 ohm) = 7 Volts
. A 0 Ω resistor is in series with a 50 Ω resistor. These are in parallel with a 80 Ω resistor. All of this is in series with a 0 Ω resistor, which is the first resistor in the circuit. The circuit is driven by a 4 V battery. This Problem comes from the Internet and you may find the solution at the following site :http://www.physics.brocku.ca/faculty/sternin/0/applets/ohm/
A. Create a table to keep track of all of the variables for this circuit then solve for all the variables. You will need to number the resistors to keep track of them and also create some other labels to keep track of everything. V I = 4 Ω V = 8 V i = A The total current for the circuit is the same as the total current in the 4 ohm resistor: Itotal = i = A The current in & 3 are the same: i = i3 The currents in 4 & 5 are the same: i4 = i5 For first junction on the diagram I = 0: IT i i4 = 0 FIND TOTAL ESISTANCE This is left to you. T = 6 = Ω V = 8/3 V i = 4/3 A 3 = Ω V3 = 4/3 V i3 = 4/3 A 4 = 5 Ω V4 = 0/3V i4 =/3 A 5 = Ω V4 = /3 V i4 =/3A T = 6 Ω VT = V IT = A FIND TOTAL CUENT V = I so I = V/, plugging in gives Itotal = (Vtotal/total) = ( V / 6 Ω ) = Amps FOM PEVIOUS EQUATION I IS KNOWN Itotal = i = A FIND VOLTAGE DOP FO V = I, plug in the appropriate values V = i = (A)(4 Ω ) = 8 V FIND VOLTAGE DOP THE PAALLEL SECTIONS The voltage drop for the parallel legs can be found using the Loop Law. This will be the voltage drop for the combination of the resistors in that leg, which are in series. (Work with the upper leg). Let VU be the Voltage drop in the upper leg and VL the Voltage drop in the lower leg: VT -V - VU = 0 so VU = VT -V = V 8 V = 4V Since the Voltage for Parallel Sections are equal VU = VL = 4V
FIND CUENT THOUGH THE PAALLEL SECTIONS Use the total resistance of each leg (upper and lower) to find the currents. The upper leg has a resistance of 3 Ω and the lower leg has a resistance of 6 Ω. Call them U = 3 Ω and L = 6 Ω. The current in either of the legs can be calculated from the total resistance of the leg and the total voltage drop of the leg as follows (using the upper leg): VU = iu U so iu = (VU / U) = 4V / 3 Ω = 4/3 Amp The equation from above: IT i i4 = 0 can be rewritten as IT iu il = 0, Solve for il gives : il = IT iu = Amp 4/3 Amp = /3 Amp Because resistors in series have the same current: il = i4 = i5 = /3 Amp iu = i = i = 4/3 Amp FIND VOLTAGE ACOSS EACH OF THE ESISTOS IN THE PAALLEL SECTIONS Use Ohm s Law: V= I so V = i = (4/3 Amp )( Ω) = 8/3 V V3 = i3 3 = (4/3 Amp )( Ω) = 4/3 V V = i4 4 = (/3 Amp )( 5 Ω) = 0/3 V V = i5 5 = (/3 Amp )( Ω) = /3 V These values may be checked with the Loop Law but this will be left to you. Enough space has been used for this problem.
Junction 3 B. This one is a bit different but only because they give you different values to start with.- The current in the 8 Ω resistor is.5 Amp. Find the current in the 0 resistor. 4 5 V I = 8 Ω V = 4 V i =.5 A = 6 Ω V = V i =.5 A 3 = 0 Ω V3 = V i3 =.75 A Notice that no single resistor has all of the current. Find the Voltage Drop for V = I so V = i = (.5 Amp)( 8 Ω) = 4 V Find the Current in Because and are in parallel they have the same Voltage Drop V = V so the currect can be calculated with Ohm s Law. 4 = 9 Ω V4 = i4 =A 5 = 8 Ω V4 = V i4 =A T =? Ω VT = V IT = A V = I so I = V/ and i = V / = (4V)/(6 Ω ) =.5 Amps Find the Current in 3 From the Current Law, the current can be calculated as follows: i + i i3 = 0 or i3 = i + i =.5 Amp +.5 Amp =.75 Amp
Given: V = 45 V P = 58 W Find: P = IV (Equation ) V= I or I =V/ (Equation ) Substitute Equation into Equation to get: V P (Equation 3) Find the total resistance in terms of : Adding 3 and 4 in series gives in parallel with and 5. Adding these in parallel gives: i i p 5 5 so this part of the circuit combines to /5. Add to this to get the total resistance of the circuit: + /5 = 7/5 total = 7/5 (Equation 4) Substitute Equation 4 into Equation 3 and solve for. V V P 5 7 P V 7 5 Substituting in the given values shows that = 5 ohms. C. The circuit in the drawing contains five identical resistors. The 45-V battery delivers 58 W to the circuit. What is the resistance of each resistor?