# Kirchhoff's Rules

## Practice

### practice problem 1

- the current through each of the connecting wires (a, b, c, d, e, f, g, h, i) with the direction (left, right);
- the voltage drop across each load element (j, k, l, m, n); and
- the resistance of each load element (j, k, l, m, n).

#### solution

- Answer
- Answer
- Answer

### practice problem 2

- the current through each of the other resistors,
- the voltage of the battery on the left, and
- the power delivered to the circuit by the battery on the right.

#### solution

- Let's identify the currents through the resistors by the value of the resistor (
*I*_{1},*I*_{2},*I*_{3},*I*_{4}) and the currents through the batteries by the side of the circuit on which they lay (*I*,_{L}*I*). Start with the 2 Ω resistor. Apply the loop rule to the circuit on the lower right._{R}20 V = *I*_{2}(2 Ω) + (3 A)(4 Ω)*I*_{2}=4 A Proceed to the 3 Ω resistor. Apply the junction rule to the junction in the center of the circuit.

*I*_{2}=*I*_{3}+*I*_{4}4 A = *I*_{3}+ 3 A*I*_{3}=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.)

*I*_{1}(1 Ω) =(4 A)(2 Ω) + (1 A)(3 Ω) *I*_{1}=11 A - 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)(1 Ω) + *V*_{L}*V*=_{L}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) + *V*_{L}*V*=_{L}9 V Good, we got the same answer using two different approaches. We must be doing the right thing.

- 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…
*I*=_{L}*I*_{1}+*I*_{3}*I*=_{L}11 A + 1 A *I*=_{L}12 A and again to the junction at the bottom…

*I*=_{R}*I*+_{L}*I*_{4}*I*=_{R}12 A + 3 A *I*=_{R}15 A to find the power of the battery on the right…

*P*=*VI**P*=(20 V)(15 A) *P*=300 W

### practice problem 3

#### solution

Let's number the currents from left to right: *I*_{1}, *I*_{2}, and *I*_{3}, respectively. Assume that the current will flow clockwise in the left circuit and counterclockwise in the right circuit; that is, that *I*_{1} and *I*_{3} are running up the page and that *I*_{2} is running down the page. Apply Kirchhoff's rules and see what happens.

I_{2} |
= I_{1} + I_{3} |
[1] top junction | |

12 V | = (4 Ω)I_{2} + (3 Ω)I_{1} |
[2] left circuit | |

5 V | = (4 Ω)I_{2} + (2 Ω)I_{3} |
[3] right circuit |

Solve using the methods of linear algebra. (We'll omit the units for clarity.)

combine [1] & [2] | |||

+4(00 | = I_{1} − I_{2} + I_{3} |
) | |

+1(12 | = 3I_{1} + 4I_{2} |
) | |

12 | = 7I_{1} + 4I_{3} |
[4] |

combine [1] & [3] | |||

+4(00 | = I_{1} − I_{2} + I_{3} |
) | |

+1(05 | = 2I_{3} + 4I_{2} |
) | |

5 | = 4I_{1} + 6I_{3} |
[5] |

combine [4] & [5] | |||

+3(12 | = 7I_{1} + 4I_{3} |
) | |

−2(05 | = 4I_{1} + 6I_{3} |
) | |

26 | = 13I_{1} |
[6] |

Continue until each current has been found.

I_{1} = +2.00 A |

I_{2} = +1.50 A |

I_{3} = −0.50 A |

The negative value of *I*_{3} means that current is running down the page, not up as we assumed. This shows the self-correcting nature of Kirchhoff's rules.

### practice problem 4

#### solution

Answer it.