Kirchhoff's Rules

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Discussion

introduction

Gustav Robert Kirchhoff (1824-1887) Prussia-Germany-Russian Empire?

Summary

Problems

practice

  1. A fairly complicated three-wire circuit is shown below [magnify]. The source voltage is 120 V between the center (neutral) and the outside (hot) wires. Load currents on the upper half of the circuit are given as 10 A, 4 A, and 8 A for the load resistors j, k, and l, respectively. Load currents on the lower half of the circuit are given as 6 A and 12 A for the load resistors m and n, respectively. The resistances of the connecting wires a, b, c, d, e, f, g, h, and i are also given.

    Determine …
    1. the current through each of the connecting wires (a, b, c, d, e, f, g, h, i) with the direction (left, right);
    2. the voltage drop across each load element (j, k, l, m, n); and
    3. the resistance of each load element (j, k, l, m, n).
    Solutions …
    1. Answer
    2. Answer
    3. Answer
  2. Given the circuit below [magnify] with 3 A of current running through the 4 Ω resistor as indicated …

    Determine …
    1. the current through each of the other resistors,
    2. the voltage of the battery on the left, and
    3. the power delivered to the circuit by the battery on the right.
    Solutions …
    1. Let's identify the currents through the resistors by the value of the resistor (I1, I2, I3, I4) and the currents through the batteries by the side of the circuit on which they lay (IL, IR). Start with the 2 Ω resistor. Apply the loop rule to the circuit on the lower right.
         
      20 V =  I2(2 Ω) + (3 A)(4 Ω)
      I2 =  4 A
         
      Proceed to the 3 Ω resistor. Apply the junction rule to the junction in the center of the circuit.
         
      I2 =  I3 + I4
      4 A =  I3 + 3 A
      I3 =  1 A
         
      The current through the 1 Ω resistor most certainly runs from right to left. If we apply the loop rule to the top circuit, we'll have to run against that current. This changes what is normally considered a potential drop into a potential increase. (Kind of like skiing up a mountain instead of down.)
         
      I1(1 Ω) =  (4 A)(2 Ω) + (1 A)(3 Ω)
      I1 =  11 A
         
    2. Apply the loop rule to the outer circuit to get the voltage of the battery on the left (continuing with the assumption that the current is running counterclockwise). We find ourselves running through the left battery backwards. This changes what is normally considered a potential increase into a potential decrease. (Kind of like using the ski lift to travel down a mountain instead of up.)
         
      20 V =  (11 A)(12 Ω) + V2
      VL =  9 V
         
      Let's verify this result by repeating the procedure for the bottom circuit.
         
      20 V =  (4 A)(2 Ω) + (1 A)(3 Ω) + V2
      VL =  9 V
         
      Good, we get the same answer by two methods. We must be doing the right thing.
    3. The power delivered to the circuit by the battery on the right is the product of its voltage times the current it drives around the circuit. We already have the voltage (it's given in the problem) all that remains is to determine the current. Apply the junction rule to the junction on the left …
         
      IL =  I1 + I3
      IL =  11 A + 1 A
      IL =  12 A
         
      and again to the junction at the bottom …
         
      IR =  IL + I4
      IR =  12 A + 3 A
      IR =  15 A
         
      to find the power of the battery on the right …
         
      P =  VI
      P =  (20 V)(15 A)
      P =  300 W
         
  3. Determine the current through each resistor in the circuit given below [magnify].

    Solution …
    • Let's number the currents from left to right: I1, I2, and I3, respectively. Assume that the current will flow clockwise in the left circuit and counterclockwise in the right circuit; that is, that I1 and I3 are running up the page and that I2 is running down the page. Apply Kirchhoff's rules and see what happens.
             
      I2  = I1 + I3   [1] top junction
       12 V  = (4 Ω)I2 + (3 Ω)I1   [2] left circuit
       5 V  = (4 Ω)I2 + (2 Ω)I3   [3] right circuit
             
      Solve using the methods of linear algebra. (We'll omit the units for clarity.)
                                 
       combine [1] & [2]     combine [1] & [3]     combine [4] & [5] 
       +4(00  = I1 − I2 + I3 )       +4(00  = I1 − I2 + I3 )       +3(12  = 7I1 + 4I3 )   
       +1(12  = 3I1 + 4I2 )       +1(05  = 2I3 + 4I2 )       −2(05  = 4I1 + 6I3 )   
        12  = 7I1 + 4I3    [4]     5  = 4I1 + 6I3    [5]     26  = 13I1    [6]
                                 
      Continue until each current has been found.
               
      I1 = +2.00 A   I2 = +1.50 A   13 = −0.50 A
               
      The negative value of I3 means that current is running down the page, not up as we assumed. This shows the self-correcting nature of Kirchhoff's rules.
  4. Write something completely different.
    • Answer it.

numerical

  
   
  1. Determine the currents in the relatively simple three-wire circuit shown to the right. Specify whether the current is flowing up or down the wire in each case.
  1. Check out the circuit shown to the right.
    1. Write Kirchhoff's equations for this circuit.
    2. Find the unknown currents I1 and I2.
    3. Find the unknown battery voltage V2.
   

algebraic

  
   
  1. The circuit on the right is made of 12 identical resistors on the edges of a cube. Determine the equivalent resistance of the cube across diagonally opposite corners in terms of the resistance R of one.
   

calculus

  
   
  1. Two batteries of emf ℰ and internal resistance r are connected in parallel to a load of resistance R as shown in the diagram to the right
    1. Write Kirchhoff's equations for this circuit.
    2. Determine the current through the load.
    3. Show that the power dissipated by the load is a maximum when R = r/2.

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