Basic Electronics Lecture 7: Resistors in Series and Parallel and Kirchhoff’s Current Law

Power Rating of Resistors in Series Circuits

• Problem Statement: In a series circuit containing four resistors ($R_1, R_2, R_3, R_4$) connected to a $120\,V$ source, determine if the indicated power rating of $\frac{1}{2}\,W$ for each resistor is sufficient to handle the actual power when the switch is closed. If the rating is not adequate, specify the minimum required rating.
• Circuit Components:     * $V_s = 120\,V$     * $R_1 = 1.0\,k\Omega$ with an indicated rating of $\frac{1}{2}\,W$     * $R_2 = 2.7\,k\Omega$ with an indicated rating of $\frac{1}{2}\,W$     * $R_3 = 910\,\Omega$ with an indicated rating of $\frac{1}{2}\,W$     * $R_4 = 3.3\,k\Omega$ with an indicated rating of $\frac{1}{2}\,W$
• Step 1: Determine Total Resistance ($R_T$):     * $R_T = R_1 + R_2 + R_3 + R_4$     * $R_T = 1.0\,k\Omega + 2.7\,k\Omega + 910\,\Omega + 3.3\,k\Omega = 7.91\,k\Omega$
• Step 2: Calculate the Circuit Current ($I$):     * $I = \frac{V_s}{R_T}$     * $I = \frac{120\,V}{7.91\,k\Omega} = 15\,mA$
• Step 3: Calculate the Power Dissipated in Each Resistor ($P = I^2 R$):     * $P_1 = I^2 R_1 = (15\,mA)^2(1.0\,k\Omega) = 225\,mW$     * $P_2 = I^2 R_2 = (15\,mA)^2(2.7\,k\Omega) = 608\,mW$     * $P_3 = I^2 R_3 = (15\,mA)^2(910\,\Omega) = 205\,mW$     * $P_4 = I^2 R_4 = (15\,mA)^2(3.3\,k\Omega) = 743\,mW$
• Evaluation of Power Ratings:     * The power rating for $R_1$ ($225\,mW$) and $R_3$ ($205\,mW$) is sufficient as they are below $\frac{1}{2}\,W$ ($500\,mW$).     * The power rating for $R_2$ ($608\,mW$) and $R_4$ ($743\,mW$) is NOT adequate because they exceed the $\frac{1}{2}\,W$ limit ($500\,mW$). For these resistors, a minimum rating higher than the calculated wattage (likely a $1\,W$ resistor) would be required.

Resistors in Parallel

• Total Resistance Formula: The reciprocal of the total resistance in a parallel circuit is the sum of the reciprocals of the individual resistances.     * $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$
• Voltage in Parallel Circuits:     * In a parallel configuration, the voltage across each resistor is equal to the applied source voltage.     * Example (Figure 9): A circuit features five resistors ($R_1=1.8\,k\Omega$, $R_2=1.0\,k\Omega$, $R_3=3.3\,k\Omega$, $R_4=2.2\,k\Omega$, $R_5=2.2\,k\Omega$) and a fuse connected in parallel to a $25\,V$ source ($V_s$).     * Voltage Findings:         * The voltage across each of the five resistors is equal to the source: $V_1 = V_2 = V_3 = V_4 = V_5 = V_s = 25\,V$.         * There is no voltage across the fuse (assuming it is ideal and not blown).

Kirchhoff’s Current Law (KCL)

• The fundamental principle of KCL: "The sum of the currents into a node (total current in) is equal to the sum of the currents out of that node (total current out)."
• Mathematical Expression:     * $I_T = I_1 + I_2$
• Example Calculations:     * Example A: If $I_1 = 5\,mA$ and $I_2 = 12\,mA$ entering/exiting a node, the total current is $I = 5\,mA + 12\,mA = 17\,mA$.     * Example B: If a circuit has a total current $I_T = 100\,mA$, and two known branch currents are $I_1 = 30\,mA$ and $I_3 = 20\,mA$, the current $I_2$ can be deduced via KCL.
• Circuit Nodes: Complex interactions across various nodes (labeled A1, A2, A3, A4, A5) demonstrate current splitting and recombining throughout a parallel resistor network using $R_1, R_2,$ and $R_3$. Values recorded include $1.5\,mA$, $1\,mA$, and $5\,mA$.

Special Cases for Parallel Resistors

• Equal-Value Resistors: When $n$ number of resistors with the same resistance value ($R$) are in parallel, the total resistance ($R_T$) is simplified to:     * $R_T = \frac{R}{n}$
• Determining an Unknown Parallel Resistor ($R_x$): If the total resistance ($R_T$) and one of the parallel resistors ($R_A$) are known, the unknown resistor ($R_x$) can be calculated using the following derivation:     * $\frac{1}{R_T} = \frac{1}{R_A} + \frac{1}{R_x}$     * $\frac{1}{R_x} = \frac{1}{R_T} - \frac{1}{R_A}$     * $\frac{1}{R_x} = \frac{R_A - R_T}{R_A R_T}$     * $R_x = \frac{R_A R_T}{R_A - R_T}$
• Current Sources in Parallel: Current sources in parallel add algebraically to find the total current ($I_{T}$).     * Example: A $50\,mA$ source and a $20\,mA$ source in parallel:     * $I_{T} = I_1 + I_2 = 50\,mA + 20\,mA = 70\,mA$

Current Dividers

• Current Divider Formula: To find the current through any specific branch ($I_x$) in a parallel circuit:     * $I_x = \left(\frac{R_T}{R_x}\right) I_T$
• Example Assessment (10 mA Total Current):     * Given resistors: $R_1 = 680\,\Omega, R_2 = 330\,\Omega, R_3 = 220\,\Omega$; total current $I_T = 10\,mA$.     * Calculate Total Resistance ($R_T$):     * $\frac{1}{R_T} = \frac{1}{680\,\Omega} + \frac{1}{330\,\Omega} + \frac{1}{220\,\Omega}$     * $R_T = 111\,\Omega$     * Calculate Individual Branch Currents ($I_x$):     * $I_1 = \left(\frac{111\,\Omega}{680\,\Omega}\right) \times 10\,mA = 1.63\,mA$     * $I_2 = \left(\frac{111\,\Omega}{330\,\Omega}\right) \times 10\,mA = 3.36\,mA$     * $I_3 = \left(\frac{111\,\Omega}{220\,\Omega}\right) \times 10\,mA = 5.05\,mA$

Power in Parallel Circuits

• Determining Total Power via Individual Resistors:     * Circuit values: $R_1 = 68\,\Omega, R_2 = 33\,\Omega, R_3 = 22\,\Omega$; $I_T = 200\,mA$.     * Step 1: Calculate Total Resistance ($R_T$):     * $\frac{1}{R_T} = \frac{1}{68\,\Omega} + \frac{1}{33\,\Omega} + \frac{1}{22\,\Omega}$     * $R_T = 11.1\,\Omega$     * Step 2: Calculate Power using Total Resistance and Current:     * $P_T = I_T^2 R_T = (200\,mA)^2(11.1\,\Omega) = 444\,mW$
• Verification Method (Individual Power Summation):     * First, determine the voltage across all branches ($V$):     * $V = I_T R_T = (200\,mA)(11.1\,\Omega) = 2.22\,V$     * Calculate power for each resistor separately using $P = \frac{V^2}{R}$, as voltage is constant across parallel branches:     * $P_1 = \frac{(2.22\,V)^2}{68\,\Omega} = 72.5\,mW$     * $P_2 = \frac{(2.22\,V)^2}{33\,\Omega} = 149\,mW$     * $P_3 = \frac{(2.22\,V)^2}{22\,\Omega} = 224\,mW$     * Summing these individual powers ($72.5\,mW + 149\,mW + 224\,mW$) yields the same total result of approximately $444\,mW$ (minor discrepancy due to rounding).