# Lab 08: Kirchhoff's Voltage Law and Kirchhoff's Current Law

## Objectives

• Verify Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) using mesh and nodal analysis of the given circuit.

## Background

#### Kirchhoff’s Voltage Law

Kirchhoff’s Voltage Law states that the algebraic sum of all the voltages around any closed path (loop or mesh) is zero: $$\sum\nolimits_i {{v_i}} = 0$$

Applying Kirchhoff’s voltage law to the first and the second loops in the circuit shown in Figure 1 yields:

Figure 1: A simple circuit to illustrate KVL and KCL.

Loop Equations:

$$\begin{array}{l} Loo{p_1}:\quad - {V_S} + {V_1} + {V_2} + {V_5} = 0 & ....(1a)\\ Loo{p_2}:\quad - {V_2} + {V_3} + {V_4} = 0 & ....(1b) \end{array}$$

$$\begin{array}{l} Loo{p_1}:\quad - {V_s} + {i_1}({R_1} + {R_2} + {R_5}) - {i_2} \cdot {R_2} = 0 & ...(2a)\\ Loo{p_2}:\quad - {i_1} \cdot {R_2} + {i_2}({R_2} + {R_3} + {R_4}) = 0 & ...(2b) \end{array}$$

#### Kirchhoff’s Current Law

Kirchhoff’s Current Law states that the algebraic sum of all the currents at any node is zero: $$\sum\nolimits_i {{i_i}} = 0$$

Applying Kirchhoff’s current law to the first four nodes in the circuit shown in Figure 1 yields the following equations:

Node Equations:

$$\begin{array}{l} Nod{e_A}:\quad - {I_S} + {I_1} = 0\\ Nod{e_B}:\quad - {I_1} + {I_2} + {I_3} = 0\\ Nod{e_C}:\quad - {I_3} + {I_4} = 0\\ Nod{e_D}:\quad - {I_2} - {I_4} + {I_S} = 0 \end{array}$$

$$Nod{e_A}:\quad \frac{{{V_A}}}{{{R_1}}} - \frac{{{V_B}}}{{{R_1}}} - \frac{{{V_D}}}{{{R_5}}} = 0$$
$$Nod{e_B}:\quad - \frac{{{V_A}}}{{{R_1}}} + {V_B}(\frac{1}{{{R_1}}} + \frac{1}{{{R_3}}} + \frac{1}{{{R_5}}}) - \frac{{{V_C}}}{{{R_3}}} - \frac{{{V_D}}}{{{R_5}}} = 0$$
$$Nod{e_C}:\quad - \frac{{{V_B}}}{{{R_3}}} + {V_C}(\frac{1}{{{R_3}}} + \frac{1}{{{R_4}}}) - \frac{{{V_D}}}{{{R_4}}} = 0$$
$$Nod{e_D}:\quad - \frac{{{V_B}}}{{{R_2}}} - \frac{{{V_C}}}{{{R_4}}} + {V_D}(\frac{1}{{{R_2}}} + \frac{1}{{{R_4}}} + \frac{1}{{{R_5}}}) = 0$$

## Questions

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