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Branch Current Method

Use both of Kirchoff’s laws. But be aware that an arbitrary application of Kirchoff’s two equations will not always yield an independent set of equations. But the following approach will probably work.

1. Label the current in each branch (do not worry about the direction of the actual current).
2. Use only interior loops and all but one node.
3. Solve the system of algebraic equations.

4. Loop Current Method

   This method is also referred to as the mesh loop method. The independent current variables are taken to be the circulating current in each of the interior loops.

   1. Label interior loop currents on a diagram.
   2. Obtain expressions for the voltage changes around each interior loop.
   3. Solve the system of algebraic equations.

   Depending on the problem, it may ultimately be necessary to algebraically sum two loop currents in order to obtain the needed interior branch current for the final answer.

   Lets consider the example of the Wheatstone bridge circuit shown in figure 1.6. We wish to calculate the currents around the loops. The three currents are identified as: the clockwise current around the large interior loop which includes the EMF, the clockwise current around the top equilateral triangle, and the clockwise current around the bottom equilateral triangle. The voltage loop expressions for the three current loops are

   
   Figure 1.6: Loop method for the Wheatstone bridge circuit.

   

   

   

   Collecting terms containing the same current gives

   

   

   

   If the values for the parameters shown in the diagram are used, the current values can be found by solving the set of simultaneous equations to give

   

   Moreover, if we number the individual currents through each resistor using the same scheme as we have for each component (current through is , has , etc.) and identify as the current out of the battery, then

   

   

   

   

   

   

   These are the same currents that would be found using only Kirchoff’s equations; however, here we had to handle only three simultaneous equations instead of six.

   Example: Use the loop current method to determine the voltage developed across the terminals AB in the circuit shown in figure 1.7.

   
   Figure 1.7: Example circuit for analysis using the loop current method.

   Consider the clockwise current loop through the two resistors and the two potentials. Similarly consider the clockwise current around the other internal loop consisting of the three resistors and . Kirchoff’s law gives

   

   

   Solving the above two equations for the unknown loop currents and gives

   

   

   

   

   The voltage across AB is given simply by

   

   Equivalent Circuits

   Equivalent circuits is often the hardest concept and most numerically intensive in the course. Learning them well could make a difference on your midterm exam. Look in several books until you find the explanation you understand best.

   Since Ohm’s law and Kirchoff’s equations are linear, we can replace any DC circuit by a simplified circuit. Just like a combination of resistors and Ohm’s law could give an equivalent resistor, a combination of circuit elements and Kirchoff’s laws can give an equivalent circuit. Two possibilities are shown in figure

   Thevenin’s and Norton’s Theorems

   A Thevenin equivalent circuit contains an equivalent voltage source in series with an equivalent resistor . A Norton equivalent circuit contains an equivalent current source in parallel with an equivalent resistor .

   
   Figure 1.8: Thevenin and Norton equivalent circuits.

   Determination of Thevenin and Norton Circuit Elements

   One approach to determine the equivalent circuits is:

   1. Thevenin – calculate the open-circuit voltage .
   2. Norton – calculate the short-circuit current between A and B; .
   3. .

   An alternative to step 2 is to short all voltage sources, open all current sources, and calculate the equivalent resistance remaining between A and B. We will use the latter approach whenever manageable. To see if you understand equivalent circuits so far, convince yourself that .

   Solution: From Thevenin’s theorem

   

   According to Notron’s theorem

   

   Therefore

   

   

   Lets now return to our Wheatstone bridge example shown in figure 1.6. We will calculate the current through by replacing the rest of the circuit by its Thevenin equivalent.

   • is removed and the open terminals are labeled . The polarity assigned is arbitrary as will be verified in the calculations.
   • The evaluation of is performed using Kirchoff’s laws:

     

     

     

     The result is V. The minus sign means only that the arbitrary choice of polarity was incorrect.

     
     Figure 1.9: Thevenin’s theorem applied to the Wheatstone bridge circuit.

   • The voltage source is shorted out and is calculated (figure 1.9):

     

     Note that when the source is shorted out, the resistors that were in series ( and ; and ) become parallel combinations.

   • The network is assembled in series as shown in figure 1.9 and the current through is calculated.

     

   Note that the numerical value of the current is the same as that in the preceding calculations, but the sign is opposite. This is simply due to the incorrect choice of polarity of for this calculation. In fact, the current flow is in the same direction in both examples, as would be expected.

   Example: Find the Thevenin equivalent components and for the circuit in figure 1.10.

   
   Figure 1.10: Example circuit for analysis using a Thevenin equivalent circuit.

   Shorting the V‘s to find gives two resistors in parallel, which are in series with a third resistor:

   

   The open circuit voltage gives . For the open circuit no current flows from the node joining the two resistors to A. A is thus at -V relative to this node. Around the interior loop (cf. voltage divider).

   Therefore