Ch 18 Direct Current Circuits concept #2, 5, 10, 12, 13, 23 Problems #1, 5, 6, 11, 17, 25, 31, 32, 33, 35, 36, 37
currents are maintained by a source of emf (battery, generator) Sources of emf act as charge pump. They increase the potential energy of the circulating charges. Emf is the work done per unit charge. SI unit is the volt. Sources of emf increase the potential of the flowing charge. When the charge moves through a resistor the potential is deceased by: I R
Batteries Perfect battery will have no resistance. Real battery has an internal resistance. The potential drop across the battery (terminal voltage) equals the battery s emf. (This is true if we neglect the internal resistance of the battery)
Battery Because a real battery has an internal resistance, the terminal voltage is not the same as the emf. As charges pass through the battery, the potential is increased by the emf,, but then is reduced by the amount I r where I is the current and r is the internal resistance. V = I r
Conclusion: is equal to the terminal voltage when the current is zero. The potential differences across the battery and the load resistor must be the same. (fig 18.1) Potential difference across resistor is V = IR Combining V = I r and V = IR We get = IR +Ir Solving for current: I = /(R+r)
The current depends on the external resistance and the internal resistance of the battery. If R is much larger then r we can ignore the internal resistance. Power = I V = I = I 2 R +I 2 r Again if R>>r then we see most of the power delivered by the battery is transferred to the resistor.
Resistors in Series Resistors connected end to end. Share 1 common point along the circuit. Currents in resistors in series are the same. Whatever comes out the first resistor goes into the second. V= IR 1 +IR 2 = I(R 1 +R 2 ) The potential difference across all the two resistors is the sum of the individual potential differences.
Resistors in series R eq is the sum of the individual resistors in series. V= IR 1 +IR 2 = I(R 1 +R 2 ) becomes V = I R eq R eq = R 1 +R 2 For more than 2 resistors R eq = R 1 + R 2 + R 3 + do page 595
Resistors in Parallel Resistors in parallel share 2 common points. See fig 18.6 Consider 2 resistors in parallel. V across each resistor is the same because they are both connected across the battery. Because charge is conserved, the current I that enters at point (A) must equal I 1 +I 2. I = I 1 +I 2
Parallel I = V/R so I 1 = V/R 1 and I 2 = V/R 2 I = V/R eq = V/R 1 + V/R 2 so: 1/R eq = 1/R 1 + 1/R 2 If more than 2 resistors are in parallel: 1/R eq = 1/R 1 + 1/R 2 + 1/R 3 + 1/R 4 +... example 18.2 quick quiz 18.3
Kirchhoff s Rules and complex DC circuit. 1) Junction rule the sum of the currents entering any junction must equal the sum of the currents leaving that junction. 2) Loop rule the sum of the potential differences across all the elements around a closed circuit loop must be zero.
Rules for circuits See page 602 work example 18.5
RC Circuits circuits with resistors and capacitors In a RC circuit, the current is not constant. RC circuits are useful in timing devices. This is because it takes some time for the capacitor to charge up. remember charge on a capacitor is Q = CV
RC circuits charging capacitors See figure 18.16 When the switch is closed, the battery begins to charge up the capacitor. As the capacitor charges the current changes. Once the capacitor is fully charged, the current becomes zero. At max charge the charge is Q = C charge on the capacitor becomes varies with time q = Q(1-e -t/rc ) voltage behaves similarly
RC = = time constant After 1 time constant the capacitor has charged up to 63.2% its maximum charge. 0.632 = 1/e After 2 time constants the capacitor charges up 63.2% of the remaining amount to the maximum charge. Capacitors charge slowly if they have a long time constant. Capacitors charge quickly if they have a short time constant. After t = 10, the capacitor is over 99.99% charged.
As the capacitor charges If we study the behavior of charging capacitors we see that after a long time the capacitor behaves like an open circuit (break in the circuit) See quick quiz 18.6
Discharging capacitors Consider a charged up capacitor. (fig. 18.17) When the switch is closed, charge begins to flow through the resistor from one capacitor plate to the other. The capacitor is discharging. IF the switch was closed at time = 0, we see that the charge on the capacitor as a function of time is: q = Q e -t/rc Q was the original charge. Can see similar function for voltage.
Do examples 18.6 and 18.7