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Chapter 18 Direct Current Circuits

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Sources of emf The source that maintains the current in a closed circuit is called a source of emf Any devices that increase the potential energy of charges circulating in circuits are sources of emf (e.g. batteries, generators, etc.) The emf is the work done per unit charge SI units: Volts

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emf and Internal Resistance A real battery has some internal resistance r ; therefore, the terminal voltage is not equal to the emf The terminal voltage: ΔV = V b – V a ΔV = ε – Ir For the entire circuit ( R – load resistance): ε = ΔV + Ir = IR + Ir

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emf and Internal Resistance ε = ΔV + Ir = IR + Ir ε is equal to the terminal voltage when the current is zero – open-circuit voltage I = ε / (R + r) The current depends on both the resistance external to the battery and the internal resistance When R >> r, r can be ignored Power relationship: I ε = I 2 R + I 2 r When R >> r, most of the power delivered by the battery is transferred to the load resistor

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Resistors in Series When two or more resistors are connected end-to-end, they are said to be in series The current is the same in all resistors because any charge that flows through one resistor flows through the other The sum of the potential differences across the resistors is equal to the total potential difference across the combination

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Resistors in Series The equivalent resistance has the effect on the circuit as the original combination of resistors (consequence of conservation of energy) For more resistors in series: The equivalent resistance of a series combination of resistors is greater than any of the individual resistors

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Resistors in Parallel The potential difference across each resistor is the same because each is connected directly across the battery terminals The current, I, that enters a point must be equal to the total current leaving that point (conservation of charge) The currents are generally not the same

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Resistors in Parallel

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For more resistors in parallel: The inverse of the equivalent resistance of two or more resistors connected in parallel is the algebraic sum of the inverses of the individual resistance The equivalent is always less than the smallest resistor in the group Resistors in Parallel

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Problem-Solving Strategy Combine all resistors in series They carry the same current The potential differences across them are not necessarily the same The resistors add directly to give the equivalent resistance of the combination: R eq = R 1 + R 2 + …

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Problem-Solving Strategy Combine all resistors in parallel The potential differences across them are the same The currents through them are not necessarily the same The equivalent resistance of a parallel combination is found through reciprocal addition:

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Problem-Solving Strategy A complicated circuit consisting of several resistors and batteries can often be reduced to a simple circuit with only one resistor Replace resistors in series or in parallel with a single resistor Sketch the new circuit after these changes have been made Continue to replace any series or parallel combinations Continue until one equivalent resistance is found

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Problem-Solving Strategy If the current in or the potential difference across a resistor in the complicated circuit is to be identified, start with the final circuit and gradually work back through the circuits (use formula ΔV = I R and the procedures describe above)

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Find the current in the 12-Ω resistor in the figure. Chapter 18 Problem 13

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Kirchhoff’s Rules There are ways in which resistors can be connected so that the circuits formed cannot be reduced to a single equivalent resistor Two rules, called Kirchhoff’s Rules can be used instead: 1) Junction Rule 2) Loop Rule Gustav Kirchhoff 1824 – 1887

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Kirchhoff’s Rules Junction Rule (A statement of Conservation of Charge): The sum of the currents entering any junction must equal the sum of the currents leaving that junction Loop Rule (A statement of Conservation of Energy): The sum of the potential differences across all the elements around any closed circuit loop must be zero

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Junction Rule I 1 = I 2 + I 3 Assign symbols and directions to the currents in all branches of the circuit If a direction is chosen incorrectly, the resulting answer will be negative, but the magnitude will be correct

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Loop Rule When applying the loop rule, choose a direction for transversing the loop Record voltage drops and rises as they occur If a resistor is transversed in the direction of the current, the potential across the resistor is – IR If a resistor is transversed in the direction opposite of the current, the potential across the resistor is +IR

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Loop Rule If a source of emf is transversed in the direction of the emf (from – to +), the change in the electric potential is +ε If a source of emf is transversed in the direction opposite of the emf (from + to -), the change in the electric potential is – ε

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Equations from Kirchhoff’s Rules Use the junction rule as often as needed, so long as, each time you write an equation, you include in it a current that has not been used in a previous junction rule equation The number of times the junction rule can be used is one fewer than the number of junction points in the circuit The loop rule can be used as often as needed so long as a new circuit element (resistor or battery) or a new current appears in each new equation You need as many independent equations as you have unknowns

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Equations from Kirchhoff’s Rules

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Problem-Solving Strategy Draw the circuit diagram and assign labels and symbols to all known and unknown quantities Assign directions to the currents Apply the junction rule to any junction in the circuit Apply the loop rule to as many loops as are needed to solve for the unknowns Solve the equations simultaneously for the unknown quantities Check your answers

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Chapter 18 Problem 23 Using Kirchhoff’s rules, (a) find the current in each resistor shown in the figure and (b) find the potential difference between points c and f.

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RC Circuits If a direct current circuit contains capacitors and resistors, the current will vary with time When the circuit is completed, the capacitor starts to charge until it reaches its maximum charge ( Q = Cε ) Once the capacitor is fully charged, the current in the circuit is zero

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Charging Capacitor in an RC Circuit The charge on the capacitor varies with time q = Q (1 – e -t/RC ) The time constant, = RC, represents the time required for the charge to increase from zero to 63.2% of its maximum In a circuit with a large (small) time constant, the capacitor charges very slowly (quickly) After t = 10 , the capacitor is over 99.99% charged

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Discharging Capacitor in an RC Circuit When a charged capacitor is placed in the circuit, it can be discharged q = Qe -t/RC The charge decreases exponentially At t = = RC, the charge decreases to 0.368 Q max ; i.e., in one time constant, the capacitor loses 63.2% of its initial charge

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Chapter 18 Problem 54 An emf of 10 V is connected to a series RC circuit consisting of a resistor of 2.0 × 10 6 Ω and a capacitor of 3.0 μF. Find the time required for the charge on the capacitor to reach 90% of its final value.

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Electrical Safety Electric shock can result in fatal burns Electric shock can cause the muscles of vital organs (such as the heart) to malfunction The degree of damage depends on –the magnitude of the current –the length of time it acts –the part of the body through which it passes

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Effects of Various Currents 5 mA or less –Can cause a sensation of shock –Generally little or no damage 10 mA –Hand muscles contract –May be unable to let go a of live wire 100 mA –If passes through the body for just a few seconds, can be fatal

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Answers to Even Numbered Problems Chapter 18: Problem 2 (a) 27 Ω (b) 0.44 A (c) 3.0 Ω, 1.3 A

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Answers to Even Numbered Problems Chapter 18: Problem 6 (a) 11.7 Ω (b) 1.00 A in the 12.0-Ω / 8.00-Ω resistors, 2.00 A in the 6.00-Ω / 4.00-Ω resistors, And 3.00 A in the 5.00-Ω resistor

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Answers to Even Numbered Problems Chapter 18: Problem 18 (a) 0.909 A (b) 1.82 V, point b at the lower potential

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Answers to Even Numbered Problems Chapter 18: Problem 22 (a) 60.0 Ω (b) 0.20 A (c) 2.4 W (d) 0.50 W

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Answers to Even Numbered Problems Chapter 18: Problem 36 587 kΩ

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Answers to Even Numbered Problems Chapter 18: Problem 38 (a) 8.0 A (b) 120 V (c) 0.80 A (d) 5.8 × 10 2 W

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