PHY 100 Series Circuits Worksheet Notes

Core Principles of Series Resistor Circuits

• Definition and Formula for Equivalent Resistance: In a series circuit, resistors are connected end-to-end such that there is only one path for the current to flow. The equivalent resistance ($R_{eq}$ or $R_T$) is calculated as the sum of the individual resistances:
      * $R_{eq} = R_1 + R_2 + R_3 + \dots$
• Current Consistency: The same current flows through all resistors in a series combination.
      * $I_T = I_1 = I_2 = I_3 = \dots$
• Voltage Distribution: The total voltage provided by the source is equal to the sum of the individual voltage drops across each resistor.
      * $V_{total} = V_1 + V_2 + V_3 + \dots$
• Ohm's Law Application: All individual calculations for voltage, current, and resistance rely on the fundamental relationship:
      * $V = I \times R$

Numerical Practice Problems (Questions 1–5)

• Q1: Finding Equivalent Resistance
      * Problem: Three resistors of $2\,\Omega$, $4\,\Omega$, and $6\,\Omega$ are connected in series. Find the equivalent resistance.
      * Given: $R_1 = 2\,\Omega$, $R_2 = 4\,\Omega$, $R_3 = 6\,\Omega$.
      * Calculation: $R_{eq} = R_1 + R_2 + R_3 = 2\,\Omega + 4\,\Omega + 6\,\Omega$
      * Result: $R_{eq} = 12\,\Omega$

• Q2: Calculating Total Current
      * Problem: A series circuit has resistors of $5\,\Omega$, $10\,\Omega$, and $15\,\Omega$ connected to a $12\,V$ battery. Find the total current in the circuit.
      * Steps:
          1. Determine equivalent resistance: $R_{eq} = 5\,\Omega + 10\,\Omega + 15\,\Omega = 30\,\Omega$.
          2. Apply Ohm's Law: $I = \frac{V}{R_{eq}} = \frac{12\,V}{30\,\Omega}$.
      * Result: $I = 0.4\,A$

• Q3: Calculating Total Voltage
      * Problem: In a series circuit, $R_1 = 3\,\Omega$, $R_2 = 7\,\Omega$, and the total current is $2\,A$. Calculate the total voltage across the combination.
      * Steps:
          1. Determine total resistance: $R_T = R_1 + R_2 = 3\,\Omega + 7\,\Omega = 10\,\Omega$.
          2. Apply Ohm's Law for total voltage: $V_T = I_T \times R_T = 2\,A \times 10\,\Omega$.
      * Result: $V_T = 20\,V$

• Q4: Voltage Drop Across Individual Resistors
      * Problem: A $9\,V$ battery is connected across three resistors in series: $2\,\Omega$, $3\,\Omega$, and $4\,\Omega$. Find the voltage drop across each resistor.
      * Steps:
          1. Total Resistance ($R_T$): $2\,\Omega + 3\,\Omega + 4\,\Omega = 9\,\Omega$.
          2. Total Current ($I_T$): $I_T = \frac{V_T}{R_T} = \frac{9\,V}{9\,\Omega} = 1\,A$.
          3. Since current is constant, $I_1 = I_2 = I_3 = 1\,A$.
          4. Voltage $V_1 = I \times R_1 = 1\,A \times 2\,\Omega = 2\,V$.
          5. Voltage $V_2 = I \times R_2 = 1\,A \times 3\,\Omega = 3\,V$.
          6. Voltage $V_3 = I \times R_3 = 1\,A \times 4\,\Omega = 4\,V$.

• Q5: Determining Current and Segmented Voltage
      * Problem: Two resistors of $12\,\Omega$ and $8\,\Omega$ are connected in series with a $10\,V$ source. Determine the current flowing through the circuit and voltage across each resistor.
      * Steps:
          1. $R_T = 12\,\Omega + 8\,\Omega = 20\,\Omega$.
          2. $I_T = \frac{10\,V}{20\,\Omega} = 0.5\,A$.
          3. $V_1 = 0.5\,A \times 12\,\Omega = 6\,V$.
          4. $V_2 = 0.5\,A \times 8\,\Omega = 4\,V$.

Numerical Practice Problems (Questions 6–10)

• Q6: Multi-Part Circuit Analysis
      * Problem: A student connects three resistors ($4\,\Omega, 6\,\Omega, 8\,\Omega$) in series to a $24\,V$ supply.
      * (a) Total Resistance: $R_T = 4 + 6 + 8 = 18\,\Omega$.
      * (b) Current in the circuit: $I = \frac{24\,V}{18\,\Omega} = 1.33\,A$.
      * (c) Voltage across each resistor:
          * $V_1 = 1.33\,A \times 4\,\Omega = 5.3\,V$
          * $V_2 = 1.33\,A \times 6\,\Omega = 7.98\,V$
          * $V_3 = 1.33\,A \times 8\,\Omega = 10.64\,V$

• Q7: Constant Current identification
      * Problem: Total resistance is $24\,\Omega$ and supply voltage is $12\,V$. What is the current flowing through each resistor?
      * Calculation: $I = \frac{12\,V}{24\,\Omega} = 0.5\,A$.
      * Note: Because it is a series circuit, current in each resistor remains the same at $0.5\,A$.

• Q8: Potential Difference Calculations
      * Problem: Three resistors in series (10 $\Omega$, 20 $\Omega$, 30 $\Omega$) with a total current of $0.5\,A$.
      * Calculations:
          * $V_1 = 0.5\,A \times 10\,\Omega = 5\,V$
          * $V_2 = 0.5\,A \times 20\,\Omega = 10\,V$
          * $V_3 = 0.5\,A \times 30\,\Omega = 15\,V$

• Q9: Comprehensive Series Analysis
      * Problem: $R_1 = 2\,\Omega$, $R_2 = 3\,\Omega$, $R_3 = 5\,\Omega$ with a $10\,V$ battery.
      * (a) Total resistance: $2 + 3 + 5 = 10\,\Omega$.
      * (b) Current: $I = \frac{10\,V}{10\,\Omega} = 1\,A$, where $I_1 = I_2 = I_3 = 1\,A$.
      * (c) Voltage drops: $V_1 = 1 \times 2 = 2\,V$, $V_2 = 1 \times 3 = 3\,V$, $V_3 = 1 \times 5 = 5\,V$.

• Q10: Circuit Verification
      * Problem: $24\,V$ battery connected to four resistors ($1\,\Omega, 2\,\Omega, 3\,\Omega, 4\,\Omega$).
      * (a) Total resistance: $1 + 2 + 3 + 4 = 10\,\Omega$.
      * (b) Current: $I = \frac{24\,V}{10\,\Omega} = 2.4\,A$.
      * (c) Voltage drops:
          * $V_1 = 2.4\,A \times 1\,\Omega = 2.4\,V$
          * $V_2 = 2.4\,A \times 2\,\Omega = 4.8\,V$
          * $V_3 = 2.4\,A \times 3\,\Omega = 7.2\,V$
          * $V_4 = 2.4\,A \times 4\,\Omega = 9.6\,V$
      * (d) Verification: $V_1 + V_2 + V_3 + V_4 = 2.4 + 4.8 + 7.2 + 9.6 = 24\,V$. This sum equals the given source voltage.

Circuit Analysis Tables and Step-by-Step Solutions

• Circuit Table 1 (3 Resistors)
      * Source Voltage ($V_T$): $12\,V$
      * Resistance Values: $R_1 = 1000\,\Omega$, $R_2 = 2000\,\Omega$, $R_3 = 3000\,\Omega$
      * Calculated Values:
          * $R_T = 1000 + 2000 + 3000 = 6000\,\Omega$
          * $I_{Total} = \frac{12\,V}{6000\,\Omega} = 0.002\,A$
          * $V_1 = 0.002\,A \times 1000\,\Omega = 2\,V$
          * $V_2 = 0.002\,A \times 2000\,\Omega = 4\,V$
          * $V_3 = 0.002\,A \times 3000\,\Omega = 6\,V$

• Circuit Table 2 (4 Resistors)
      * Source Voltage ($V_T$): $10\,V$
      * Resistance Values: $R_1 = 200\,\Omega, R_2 = 300\,\Omega, R_3 = 500\,\Omega, R_4 = 1000\,\Omega$
      * Calculated Values:
          * $R_T = 200 + 300 + 500 + 1000 = 2000\,\Omega$
          * $I_{Total} = \frac{10\,V}{2000\,\Omega} = 0.005\,A$
          * $V_1 = 0.005\,A \times 200\,\Omega = 1\,V$
          * $V_2 = 0.005\,A \times 300\,\Omega = 1.5\,V$
          * $V_3 = 0.005\,A \times 500\,\Omega = 2.5\,V$
          * $V_4 = 0.005\,A \times 1000\,\Omega = 5\,V$

• Circuit Table 3 (Reverse Solve for Total Voltage)
      * Total Current ($I_T$): $50\,mA = 0.05\,A$
      * Individual Voltage Drops Given: $V_1 = 5\,V, V_2 = 3\,V, V_3 = 1\,V$
      * Calculated Values:
          * $V_{Total} = 5 + 3 + 1 = 9\,V$
          * $R_1 = \frac{5\,V}{0.05\,A} = 100\,\Omega$
          * $R_2 = \frac{3\,V}{0.05\,A} = 60\,\Omega$
          * $R_3 = \frac{1\,V}{0.05\,A} = 20\,\Omega$
          * $R_T = 100 + 60 + 20 = 180\,\Omega$

Power and Resistance in Series Circuits

• Power Formula: $P = I \times V$ or $P = I^2 \times R$
• Comprehensive Table Task:
      * Given: $V_T = 12\,V$. Resistors $R_1 = 150\,\Omega, R_2 = 220\,\Omega, R_3 = 470\,\Omega$.
      * Resistor 1 Analysis:
          * $R = 150\,\Omega$
          * $V = 2.1\,V$
          * $I = 0.014\,A$
          * $P = 0.014 \times 2.1 = 0.03\,W$
      * Resistor 2 Analysis:
          * $R = 220\,\Omega$
          * $V = 3.08\,V$
          * $I = 0.014\,A$
          * $P = 0.014 \times 3.08 = 0.04\,W$
      * Resistor 3 Analysis:
          * $R = 470\,\Omega$
          * $V = 6.58\,V$
          * $I = 0.014\,A$
          * $P = 0.014 \times 6.58 = 0.09\,W$
      * Total Circuit Analysis:
          * $R_T = 150 + 220 + 470 = 840\,\Omega$
          * $V_T = 12\,V$
          * $I_T = \frac{12}{840} = 0.014\,A$
          * $P_T = I_T \times V_T = 0.014 \times 12 = 0.17\,W$