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.

Equipments

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:

00 SampleCircuit
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

Procedure

Questions