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= –Ir
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 = I2R +I2r
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= IR1 +IR2 = I(R1 +R2)
The potential difference across all the two
resistors is the sum of the individual potential
differences.
Resistors in series
Req is the sum of the individual resistors in series.
V= IR1 +IR2 = I(R1 +R2)
becomes V = I Req Req = R1+R2
For more than 2 resistors
Req = R1 + R2 + R3 + …
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 I1 +I2.
I = I1 +I2
Parallel
I = V/R
so I1 = V/R1
and I2 = V/R2
I = V/Req = V/R1 + V/R2
so: 1/Req = 1/R1 + 1/R2
If more than 2 resistors are in parallel:
1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4 + ...
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