Lecture 2: Ohm's Law and Kirchhoff's Laws

Resistance

• Resistance is the property of a material to resist the flow of electric charge.
• Represented by the symbol $R$.

Ohm's Law

• Ohm’s law states that the voltage $v$ across a resistor is directly proportional to the current $i$ flowing through the resistor.

Special Cases

• Short Circuit:
  • A circuit element with resistance approaching zero.
  • $R = 0$, $v = 0$
• Open Circuit:
  • A circuit element with resistance approaching infinity.
  • $R = \infty$, $i = 0$

Kirchhoff's Laws

• Based on the law of conservation of charge, which requires that the algebraic sum of charges within a system cannot change.

Kirchhoff's Current Law (KCL)

• The algebraic sum of currents entering a node (or a closed boundary) is zero.
  • $\sum I<em>{in} = \sum I</em>{out}$
  • Example:
    • $I<em>T = I</em>1 - I<em>2 + I</em>3$
    • $I<em>T + I</em>2 = I<em>1 + I</em>3$
    • $I<em>1 = I</em>T - I<em>2 + I</em>3$

Kirchhoff’s Voltage Law (KVL)

• States that the algebraic sum of all voltages around a closed path (or loop) is zero.
  • $\sum V<em>{drops} = \sum V</em>{rises}$
  • Example:
    • $-V<em>{ab} + V</em>1 + V<em>2 - V</em>3 = 0$
    • $V<em>s = V</em>1 + V<em>2 - V</em>3$
    • $V<em>{ab} = V</em>1 + V<em>2 - V</em>3$

Example 1 (KVL Application)

• Problem: Determine $v_o$ and $i$ in the circuit.
• Given:
  • Voltage source: 12 V
  • Resistors: 4Ω, 6Ω
  • Voltage source: 4 V
• Solution:
  • Applying KVL around the loop:
    • $-12 + 4i + 2v_o - 4 + 6i = 0$
  • Given: $v_o = -6i$
  • Substitute $v_o$:
    • $-16 + 10i - 12i = 0$
    • $-16 - 2i = 0$
  • Solve for $i$:
    • $i = -8 A$
  • Solve for $v_o$:
    • $v_o = -6i = 48 V$

Example 2 (KCL Application)

• Problem: Find current $i<em>o$ and voltage $v</em>o$ in the circuit.

• Given:

  • Current source: 3A
  • Resistors: 4Ω

• Solution:

  • Applying KCL to node a:
    • $3 + 0.5i<em>o = i</em>o$
    • $3 = 0.5i_o$
    • $i_o = 6 A$
  • For the 4Ω resistor, Ohm's law gives:
    • $v<em>o = 4i</em>o = 24 V$

Example 3

• Problem: Find current $i<em>o$ and voltage $v</em>o$ in the circuit.
• Solution:
  • KCL at node (a): $i<em>1 + i</em>o = 6$
  • $i<em>1 = 6 - i</em>o$
  • KVL at loop (1):
    • $4i<em>1 - 2i</em>o + 8i_1 = 0$
    • From (1), one can get
      • $-2i<em>o + 4(6 - i</em>o) + 8(6 - i_o) = 0$
      • $-2i<em>o + 24 - 4i</em>o + 48 - 8i_o = 0$
      • $-14i_o + 72 = 0$
      • $i_o = \frac{72}{14} = \frac{36}{7} \approx 5.14 A$
  • $v<em>o = 8i</em>1$
  • $v<em>o = 8(6 - i</em>o) = 8(6 - \frac{36}{7}) = 8(\frac{42 - 36}{7}) = 8(\frac{6}{7}) = \frac{48}{7} \approx 6.86 V$

Example 4

• Problem: Given $I<em>o = 2 mA$ in the network, find $V</em>A$.

• Solution:

  • KVL at loop (1): $-1I<em>o - 6 + 2i</em>2 = 0$

    • $- (1)(2) - 6 + 2i_2 = 0$
    • $i_2 = 4 mA$

  • KCL at node (b): $I<em>o + i</em>2 + i_3 = 0$

    • $2 + 4 + i_3 = 0$
    • $i_3 = -6 mA$

  • KVL at loop (2): $-2i<em>2 + (1)i</em>4 + 2i_3 = 0$

    • $-2(4) + (1)i_4 + 2(-6) = 0$
    • $i_4 = 20 mA$

  • KCL at node (e): $i<em>4 - i</em>3 - i_5 = 0$

    • $20 + 6 - i_5 = 0$
    • $i_5 = 26 mA$

  • $-V<em>A - 2i</em>3 - 1i_4 = 0$

  • $-V<em>A - (1)i</em>5 - 1i_4 = 0$

    • $V<em>A = -i</em>5 - i_4 = -26 - 20 = -46 V$

Series Resistors and Voltage Division

• In series connections, the current is the same through all resistors: $i<em>s = i</em>1 = i<em>2 = i</em>3$
• The sum of voltage drops across the resistors equals the source voltage: $-V<em>s + V</em>1 + V<em>2 + V</em>3 = 0$
• $V<em>s = R</em>1i<em>1 + R</em>2i<em>1 + R</em>3i_1$
• The current $i_1$ can be calculated as:
  • $i<em>1 = \frac{V</em>s}{R<em>1 + R</em>2 + R_3}$
• The voltage across the $n^{th}$ resistor $R_n$ is:
  • $V<em>n = i</em>1R<em>n = \frac{V</em>sR<em>n}{R</em>1 + R<em>2 + R</em>3}$
• For a series connection of N resistors, the equivalent resistance is:
  • $R<em>s = R</em>1 + R<em>2 + … + R</em>N$

Parallel Resistors and Current Division

• In parallel connections, the total current $i(t)$ is the sum of the currents through each resistor: $i(t) = i<em>1(t) + i</em>2(t) + i_3(t)$
• The voltage across each parallel resistor is the same: $v(t) = R<em>1i</em>1(t) = R<em>2i</em>2(t)$
• $i(t) = \frac{v(t)}{R<em>1} + \frac{v(t)}{R</em>2} + … + \frac{v(t)}{R<em>N} = v(t)(\frac{1}{R</em>1} + \frac{1}{R<em>2} + … + \frac{1}{R</em>N})$
• Equivalent resistance $R_P$ for parallel resistors:
  • $\frac{1}{R<em>P} = \frac{1}{R</em>1} + \frac{1}{R_2}$
  • $R<em>P = \frac{R</em>1R<em>2}{R</em>1 + R_2}$
• Special Cases:
  • If resistors are equal (R) with n resistors: $R_{eq} = \frac{R}{n}$.
  • For two equal parallel resistors: $R_{eq} = \frac{R}{2}$, and the current is divided equally among them.
• The sum of conductance is:
  • $\frac{1}{R<em>p} = \sum</em>{i=1}^{N} \frac{1}{R_i}$

Example 5

• Problem: Determine the resistance at the terminals A-B in the network.
• Solution:
  • Combine resistors in series and parallel to simplify the network step-by-step.
• $R_{AB} = 6 k\Omega$

Example 6

• Problem: Determine the resistance at the terminals A-B in the network.
• Answer: $R_{AB} = 3 k\Omega$

Example 7

• Problem: We wish to find all the currents and voltages labeled in the ladder network shown.

• Solution:

  • $I_1 = 1 mA$

  • $V_a = 3 V$

  • $I_2 = 3.33 * 10^{-4} mA = \frac{1}{3} mA$

  • $I_3 = \frac{2}{3}* 10^{-3} = \frac{2}{3} mA$

  • $I<em>3 = I</em>4 + I_5$

  • $I_4 = \frac{1}{8} mA$

  • $I_5 = \frac{3}{80} mA$

  • $V_c = \frac{3}{8} V$

Example 8

• Problem: Find $V<em>o$, $V</em>1$, and $V_2$ in the circuit shown in this figure.