Week 12 Pre-work PT2 DC Circuits

EMF and Terminal Voltage

• Electric circuits require a battery or generator to produce current; these are called sources of electromotive force (EMF).
• A battery is a nearly constant voltage source but has a small internal resistance ($r$), which reduces the actual voltage from the ideal EMF ($\xi$).
• The terminal voltage ($V<em>{ab}$) is given by: $V</em>{ab} = \xi - Ir$.
• The internal resistance behaves as though it were in series with the EMF.

Example 26-1: Battery with Internal Resistance

• A $65.0 \Omega$ resistor is connected to a battery with EMF $12.0 V$ and internal resistance $0.5 \Omega$.
  • (a) Calculate the current in the circuit.
  • (b) Calculate the terminal voltage of the battery, $V_{ab}$.
  • (c) Calculate the power dissipated in the resistor $R$ and in the battery’s internal resistance $r$.

Solution to Example 26-1

• (a) Current in the circuit:
  • Total resistance: $R_{TOT} = R + r$.
  • $E = IR<em>{TOT} \rightarrow I = \frac{E}{R</em>{TOT}} = \frac{12.0}{65.0 + 0.5} = 0.183 A = 183 mA$
• (b) Terminal voltage of the battery:
  • $V_{ab} = E - Ir = 12.0 - 0.183 \times 0.5 = 11.9 V$
• (c) Power dissipated:
  • In the resistor $R$: $P = I^2 R = 0.183^2 \times 65.0 = 2.18 W$
  • In the internal resistance $r$: $P = I^2 r = 0.183^2 \times 0.5 = 0.02 W$

Resistors in Series

• A series connection has a single path from the battery, through each circuit element in turn, and back to the battery.
• The current through each resistor is the same; the voltage depends on the resistance.
• The sum of the voltage drops across the resistors equals the battery voltage: $V = V<em>1 + V</em>2 + V<em>3 = IR</em>1 + IR<em>2 + IR</em>3$.
• Equivalent resistance ($R_{eq}$) for series resistors:
  • $R<em>{eq} = \frac{V}{I} = R</em>1 + R<em>2 + R</em>3$

Resistors in Parallel

• A parallel connection splits the current; the voltage across each resistor is the same.
• The total current is the sum of the currents across each resistor:$I=I<em>1+I</em>2+I_3$
• The reciprocal of the equivalent resistance:
  $\frac{1}{R<em>{eq}} = \frac{1}{R</em>1} + \frac{1}{R<em>2} + \frac{1}{R</em>3} + …$

Conceptual Example 26-2: Series or Parallel?

• (a) Identical light bulbs in series vs. parallel.
  • Series: $R<em>{eq} = R + R = 2R$; $P</em>{series} = \frac{V^2}{R_{eq}} = \frac{V^2}{2R}$
  • Parallel: $\frac{1}{R<em>{eq}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \rightarrow R</em>{eq} = \frac{R}{2}$; $P<em>{parallel} = \frac{V^2}{R</em>{eq}} = \frac{2V^2}{R}$
  • $\frac{P<em>{parallel}}{P</em>{series}} = \frac{2V^2}{R} \times \frac{2R}{V^2} = 4$; Parallel connection is brighter.
• (b) Headlights of a car are wired in parallel so that if one burns out, the other stays on.

Conceptual Example 26-3: An Illuminating Surprise

• A $100 W$, $120 V$ lightbulb and a $60 W$, $120 V$ lightbulb connected in series and parallel configurations.
  • In parallel, each bulb sees the full $120 V$ drop, as they are designed to do, so the $100 W$ bulb is brighter.
  • In series, $P = \frac{V^2}{R}$, so at constant voltage, the bulb dissipating more power will have lower resistance. The $60 W$ bulb, whose resistance is higher, will be brighter because more of the voltage will drop across it than across the $100 W$ bulb.

Example 26-4: Circuit with Series and Parallel Resistors

• Calculate the amount of current drawn from the battery.
• $R<em>{bc} = \frac{R</em>1 R<em>2}{R</em>1 + R_2} = \frac{500 \times 700}{500 + 700} = 292 \Omega$
• $R<em>{ac} = R</em>{ab} + R_{bc} = 400 + 292 = 692 \Omega$
• $I = \frac{V}{R_{ac}} = \frac{12.0}{692} = 0.0173 A = 17.3 mA$

Example 26-5: Current in One Branch

• Determine the current through the $500 \Omega$ resistor.
• From the previous example: $I = 17.3 mA$ and $R_{bc} = 292 \Omega$.
• $V<em>{bc} = IR</em>{bc} = 0.0173 \times 292 = 5.0 V$
• $I<em>1 = \frac{V</em>{bc}}{R} = \frac{5.0}{500} = 0.010 A = 10 mA$

Conceptual Example 26-6: Bulb Brightness in a Circuit

• Three identical lightbulbs (A, B, C), each with resistance $R$.
  • (a) When switch S is closed, bulbs A and B are in parallel, and this combination is in series with bulb C. Bulbs A and B will be equally bright but much dimmer than C.
  • (b) With switch S open, no current flows through A, so it is dark. B and C are now equally bright, and each has half the voltage across it, so C is somewhat dimmer than it was with the switch closed, and B is brighter.

Example 26-8: Analyzing a Circuit

• A $9.0 V$ battery with internal resistance $r = 0.50 \Omega$ is connected in a circuit.
  • (a) Calculate how much current is drawn from the battery.
  • (b) Determine the terminal voltage of the battery.
  • (c) Find the current in the $6.0 \Omega$ resistor.

Solution to Example 26-8

• (a) Current drawn from the battery:
  • By replacing series & parallel combinations
  • $I = \frac{E}{R<em>{eq}} = \frac{E}{r + R</em>{eq3+5.0}} = \frac{9.0}{0.50 + 4.8 + 5.0} = 0.87 A$
• (b) Terminal voltage of the battery:
  • $V_T = E - Ir = 9.0 - 0.87 \times 0.50 = 8.6 V$
• **(c) Current in the $6.0 \Omega$ resistor:
• Total Voltage shared by all resistors
  • Voltage across $R<em>{eq3} = \frac{R</em>{eq3} \times E}{R_{TOT}} = \frac{4.8 \times 9.0}{4.8 + 5.0 + 0.5} = 4.19 V$
  • Voltage across $R_{eq2} = 4.19 V$
  • Voltage across $6 \Omega = \frac{6.0 \times 4.19}{6.0 + 2.7} = 2.89 V$
  • $I_{6\Omega} = \frac{V}{R} = \frac{2.89}{6.0} = 0.48 A$

Summary

• A source of EMF transforms energy from some other form to electrical energy.
• A battery is a source of EMF in parallel with an internal resistance.
• Resistors in series: $R<em>{eq} = R</em>1 + R<em>2 + R</em>3 + …$
• Resistors in parallel: $\frac{1}{R<em>{eq}} = \frac{1}{R</em>1} + \frac{1}{R<em>2} + \frac{1}{R</em>3} + …$