Circuits

Overview of Current Through Resistors

• Current in Series Circuits

  • The current through each resistor in a series circuit is the same.
  • Each resistor experiences the same current, denoted as I.

• Potential Difference Across Resistors

  • To find the potential difference across each resistor, use the formula:
  • $ext{Potential Difference} (V) = -IR$

Overview of Parallel Circuits

• Current Splitting at Junctions

  • In a parallel circuit, the current can divide among the different resistors.
  • The total current supplied by the battery (denoted as $I<em>0$) equals the sum of the currents through the individual resistors $I</em>1$, $I<em>2$, and $I</em>3$:
  • $I<em>0 = I</em>1 + I<em>2 + I</em>3$

• Potential Difference Across Parallel Resistors

  • The potential difference across each resistor in parallel is the same:
  • $ext{Voltage Drop} (V) = ext{Voltage Battery} (V_{ ext{battery}})$

Series and Parallel Resistance Calculations

• Calculating Equivalent Resistance for Series Resistors

  • For resistors in series, the equivalent resistance ($R_{eq}$) is given by:
  • $R<em>{eq} = R</em>1 + R<em>2 + R</em>3$
  • Example calculation:
  • If $R<em>1 = 250 ext{ ohms}, R</em>2 = 150 ext{ ohms}, R_3 = 100 ext{ ohms}$, then:
  • $R_{eq} = 250 + 150 + 100 = 500 ext{ ohms}$

• Calculating Equivalent Resistance for Parallel Resistors

  • For resistors in parallel, the equivalent resistance ($R_{eq}$) is given by:
  • rac{1}{R{eq}} = rac{1}{R1} + rac{1}{R2} + rac{1}{R3}
  • Rearranging gives us:
  • R{eq} = rac{1}{rac{1}{R1} + rac{1}{R2} + rac{1}{R3}}
  • Example calculation:
  • For $R<em>1 = 250 ext{ ohms}, R</em>2 = 150 ext{ ohms}, R_3 = 350 ext{ ohms}$, we find:
  • $R_{eq} = 73.9 ext{ ohms}$

Example Problem - Series Circuit

• Given:

  • Battery voltage: $V = 24 ext{ volts}$
  • Equivalent resistance: $R_{eq} = 500 ext{ ohms}$

• To Find:

  • Current supplied by the battery using Ohm's Law:
  • Ohm's Law formula: $ext{Voltage} (V) = Current (I) imes Resistance (R)$
  • Rearranging gives:
  • I = rac{V}{R_{eq}} = rac{24 ext{ volts}}{500 ext{ ohms}} = 0.048 ext{ A}

• Determine Potential Differences Across Each Resistor:

  • Using $ext{Potential Difference} (V) = -IR$ for each resistor:
  • $ext{For } R<em>1: V</em>1 = -I imes R_1 = -0.048 imes 250 = -12 ext{ volts}$
  • $ext{For } R<em>2: V</em>2 = -I imes R_2 = -0.048 imes 150 = -7.2 ext{ volts}$
  • $ext{For } R<em>3: V</em>3 = -I imes R_3 = -0.048 imes 100 = -4.8 ext{ volts}$

• Verification of Work:

  • Apply Kirchhoff’s Law:
  • Check if $ext{Total Voltage} = V<em>{ ext{battery}} + V</em>1 + V<em>2 + V</em>3 = 0$
  • Verify that $24 + (-12) + (-7.2) + (-4.8) = 0$

Example Problem - Parallel Circuit

• Given:

  • Battery voltage: $V = 24 ext{ volts}$
  • Resistor values:
  • $R<em>1 = 250 ext{ ohms}, R</em>2 = 150 ext{ ohms}, R_3 = 350 ext{ ohms}$
  • Calculate $R_{eq}$:
  • rac{1}{R_{eq}} = rac{1}{250} + rac{1}{150} + rac{1}{350}
  • Result: $R_{eq} = 73.9 ext{ ohms}$

• Current Supplied by Battery:

  • Using Ohm's Law:
  • I_0 = rac{24}{73.9} = 0.325 ext{ A}

• Finding Current Through Each Resistor:

  • Since $ext{Voltage across each resistor} = V_{ ext{battery}} = 24 ext{ volts}$:
  • For $R_1$:
    • I_1 = rac{24}{250} = 0.096 ext{ A}
  • For $R_2$:
    • I_2 = rac{24}{150} = 0.16 ext{ A}
  • For $R_3$:
    • I_3 = rac{24}{350} = 0.069 ext{ A}

• Verification of Work:

  • Check if the total current adds up:
  • $I<em>0 = I</em>1 + I<em>2 + I</em>3$
  • Verify that $0.325 = 0.096 + 0.16 + 0.069$

Applications of Series and Parallel Circuits

• Real World Example:
  • Christmas lights are often wired in parallel to ensure that if one bulb fails, others remain lit.
  • Historically, if wired in series, the failure of one bulb would cause the whole string to go out.

Conclusion

• Emphasize the importance of understanding series and parallel circuits for analyzing electrical systems.
• Encourage students to practice calculating equivalent resistances, current, and potential differences in different configurations.