PHYS345 Electricity and Electronics

Multiloop Circuit Example

Consider the circuit shown in A and find the

  1. Current through the batteries?
  2. Current through the 8k resistor?
  3. Voltage difference across the 20k resistor?
  4. Rate of energy dissipated by the 6k resistor?
Please note that k stands for kilo-ohms throughout this page.

I have changed the circuit parameters from the diagram shown in class. The same elements are used as the Detailed Simple Circuit Example, with the addition of one 9 V battery as shown.

Circuit schematic involving 5 resistors and 2 emfs

Step 1: Replace any combination of resistors in series or parallel with their equivalent resistance. Here I simply use the result from the previous example.

Three resistors replaced with equivalent 12k

Redraw slightly to show any symmetry in the circuit.

Symmetrized schematic

Label resistors and emfs with symbols.

Schematic with circuit elements labeled

Eliminate some of the clutter. We will not need the circuit parameters until later...

Schematic redrawn to rely on labels rather than values

Step 2: Choose a direction for the current in each branch of the circuit, and label the currents in a circuit diagram. Add plus and minus signs to indicate the high and low voltage sides of each source of emf and resistor.

Current directions selected and labeled

Step 3: Apply the junction rule to each junction where the current divides. Consider point A.

Application of the junction rule

Result of junction rule applied at point A

Step 4: In a circuit containing n interior loops, apply the rule to any n loops. First the left inner loop, clockwise from point A.

Left inner loop

First application of loop rule

Second the right inner loop, clockwise from point A.

Right inner loop

First application of loop rule

Step 5: Solve the equations to obtain the values of the unknowns.

Solve these simultaneous equations using your preferred approach

The physics is now complete and the mathematics begins. Tackle the the solution of these three simultaneous equations in your favorite way. When convenient I like to use a symbolic manipulation program to check my answers.

i1 :  4/3 mA 1.33 mA
i2 :  - 7/18 mA   - 0.39 mA
i3 :  17/18 mA 0.94 mA

Note that the current in the center branch was chosen in the wrong direction, given the negative sign of our answer.

To find the voltage difference across the 20k resistor, we have 0.94 mA flowing through the equivalent resistance of 4 k, representing the parallel combination of the 20 k and 5 k resistors. So there is a voltage difference of 3.76 V across the 20 k resistor.

The voltage difference across the 6k resistor is 0.39 mA x 6k = 2.34 V. So the power dissipated by it is 0.39 mA x 2.34 V = 0.91 mW.

In summary,
1. Current through the original batttery? 1.33 ma (was 1.00 mA in previous circuit)
2. Current through the 8k resistor? 0.94 mA (was 0.33 mA in previous circuit)
3. Voltage difference across the 20k resistor? 3.76 V (was 1.33 V in previous circuit)
4. Rate of energy dissipated by the 6k resistor? 0.91 mW (was 2.67 mW in previous circuit)

Step 6: Check your results by assigning a voltage of zero to one point in the circuit (the ground) and use the values of the currents found to determine the voltage at other points in the circuit.

The outer loop

Final application of the loop rule to check the solution

9.0 V - (1.33 mA)(5k) - (0.64 mA)(12k) + 9.0 V = 0

OK, it checks out numerically to within two significant digits.

Last updated Sept. 7, 1998.
Copyright George Watson, Univ. of Delaware, 1998.