- Current through the batteries?
- Current through the 8k resistor?
- Voltage difference across the 20k resistor?
- Rate of energy dissipated by the 6k resistor?

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.

**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.

Redraw slightly to show any symmetry in the circuit.

Label resistors and emfs with symbols.

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

**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 potential sides of each
source of emf and resistor.

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

Second the right inner loop, clockwise from point A.

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

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.

i_{1} : |
4/3 mA | 1.33 mA |

i_{2} : |
- 7/18 mA | - 0.39 mA |

i_{3} : |
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) |

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

"http://www.physics.udel.edu/~watson/phys208/circuit-example.html"

Last updated Sept. 17, 1997.

Copyright George Watson, Univ. of Delaware, 1997.