DC Circuits Series and parallel rules for resistors

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DC Circuits • Series and parallel rules for resistors • Kirchhoff’s circuit rules

DC Circuits • Series and parallel rules for resistors • Kirchhoff’s circuit rules

“DC” Circuits “Direct Current or DC”: current always flows in one direction. For circuits

“DC” Circuits “Direct Current or DC”: current always flows in one direction. For circuits containing only resistors and emf’s the current is always constant in time. Circuits containing other elements such as capacitors and inductors as well as resistors will have currents that change with time. Alternating current or AC is current that reverses direction many times (eg: 60 Hz current in Canada) and will not be treated in this course

Resistors in Series I A R 1 R 2 R 3 B We want

Resistors in Series I A R 1 R 2 R 3 B We want to replace this combination by a single resistor with resistance Reff A I B Reff

 • Same current through all resistors • Voltages add: Veff = V 1

• Same current through all resistors • Voltages add: Veff = V 1 + V 2 + V 3 + … IReff = IR 1 + IR 2 + IR 3 + … (same current through all) So, Easy way to remember: think of the length of a string: Ltot = L 1 + L 2 + L 3 where L 1 etc are the segments

Resistors in Parallel I A I 1 I 2 R 2 I 3 R

Resistors in Parallel I A I 1 I 2 R 2 I 3 R 3 I B R 1 We want to replace these resistors by a single resistance Reff: A B I Reff

 • Same voltage across each resistor • Currents add: Ieff = I 1

• Same voltage across each resistor • Currents add: Ieff = I 1 + I 2 + I 3 +

Example 1 R 1 11 V R 4 R 3 R 2 R 5

Example 1 R 1 11 V R 4 R 3 R 2 R 5 All resistors = 1 Ω Find: Effective resistance across the battery R 6

Quiz Find the effective resistance of a network of identical resistors: A) 11/6 R

Quiz Find the effective resistance of a network of identical resistors: A) 11/6 R B) 6 R C) 6/11 R D) 1/6 R R R R

Example 2 Find a) the current in each resistor b) the power dissipated by

Example 2 Find a) the current in each resistor b) the power dissipated by each resistor a 18 V c) the equivalent resistance of the three resistors 3Ω 6Ω b 9Ω

Example 3 A regular “ 40 watt” bulb and a “ 60 watt” bulb

Example 3 A regular “ 40 watt” bulb and a “ 60 watt” bulb are connected in SERIES across 120 V. What power does each bulb give? (Assume that the resistances don’t change with temperature—these are special bulbs. )

Kirchhoff’s Circuit Rules Junction Rule: total current in = total current out at each

Kirchhoff’s Circuit Rules Junction Rule: total current in = total current out at each junction (from conservation of charge). Loop Rule: Sum of potential differences around any closed loop is zero (from conservation of energy). Charge q moves through circuit changing its potential energy q. V but eventually there is no overall change.

Junction Rule: conservation of charge. I 1 I 2 I 1 = I 2

Junction Rule: conservation of charge. I 1 I 2 I 1 = I 2 + I 3 Sum of currents entering a junction equals the sum of currents leaving the junction Alternately: I 1 I 2 I 1 = I 2 + (I 1 -I 2) I 1 -I 2

Loop Rule: conservation of energy. Follow a test charge q around a loop: around

Loop Rule: conservation of energy. Follow a test charge q around a loop: around any loop in circuit. R I - + -Q +Q ΔV = -IR ΔV = Q/C C loop going from left to right

Example 5 A dead battery is charged by connecting it to the live battery

Example 5 A dead battery is charged by connecting it to the live battery of another car with jumper cables. Determine the current in the starter and in the dead battery.

Solution

Solution

Example In the circuit shown, currents of 2. 50 A and 1. 60 A

Example In the circuit shown, currents of 2. 50 A and 1. 60 A flow in the branches indicated in the diagram. Apply Kirchhoff’s circuit rules and calculate the potential difference Ve Vb between points b and e in the diagram. 2. 5 A 2. 0 c 3. 0 d I 2 I 1 R e b 4. 0 2. 5 A I 3 a 1. 6 A 6. 0 f