1. Direct-Current Circuits

2. Resistors in Series and Parallel
As with capacitors, resistors are often in series and parallel configurations in circuits
Series
Parallel
The question then is what is the equivalent resistance

3. Resistors in Series
Since these resistors are in series, we have the same current in all three resistors
We also have that the sum of the potential differences across the three resistors must be the same as the potential difference between points a and b

4. Resistors in Series Then using We have that
Now the equivalent resistor, R, will also have the same potential difference across it as Vab, and it will also have the same current I
Equating these last two results, we then have that
The equivalent resistance for a sequence of resistors in series is just the sum of the individual resistances

5. Resistors in Parallel
Here we have that the voltage across each resistor has to be the same (work done in going from a to b is independent of the path, independent of which resistor you go through)

6. Resistors in Parallel We now deal with currents through the resistors
At point a the current splits up into three distinct currents
We have that the sum of theses three currents must add to the value coming into this point
We also have that
The equivalent resistor, R, will have also have the current I going through it

7. Resistors in Parallel Using
and combining with the previous equations, we then have or
The inverse of the effective resistance is given by the sum of the inverses of the individual resistances

8. Solving Resistor Networks
Make a drawing of the resistor network
Determine whether the resistors are in series or parallel or some combination
Determine what is being asked
Equivalent resistance
Potential difference across a particular resistance
Current through a particular resistor

9. Solving Resistor Networks
Solve simplest parts of the network first
Then redraw network using the just calculated effective resistance
Repeat calculating effective resistances until only one effective resistance is left

10. Solving Resistor Networks
Given the following circuit
What is the equivalent resistance and what is the current through each resistor
We see that we have two resistors in parallel with each other and the effective resistance of these two is in series with the remaining resistor

11. Solving Resistor Networks
Step 1: Combine the two resistors that are in parallel
yielding
Step 2: Combine the two resistors that are in series
yielding

12. Solving Resistor Networks
Current through this effective resistor is given by
The current through the resistors in the intermediate circuit of Step 1 is also 3 Amps with the voltage drop across the individual resistors being given by

13. Solving Resistor Networks
To find the current through the resistors of the parallel section of the initial circuit, we use the fact that both resistors have the same voltage drop – 6 Volts

14. Consistency Check
There is a check that can be made to see if the answers for the currents make sense:
The power supplied by the battery should equal the total power being dissipated by the resistors
The power being supplied by the battery is given by
where I is the total current
The power being dissipated by each of the resistors is given by

15. Example 1
Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open.
If switch S is closed, what happens to the brightness of the bulb R2?
a) It increases b) It decreases c) It doesn’t change
The power dissipated in R2 is given by
When the switch is closed neither V nor R changes
So the brightness does not change

16. Example 2
Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open.
What happens to the current I, after the switch is closed ?
a) Iafter = 1/2 Ibefore b) Iafter = Ibefore c) Iafter = 2 Ibefore
Initially the current is given by
After the switch is closed the net resistance is given by
The new current is then

17. Kirchoff’s Rules Not all circuits are reducible
There is no way to reduce the four resistors to one effective resistance or to combine the three voltage sources to one voltage source

18. Kirchoff’s Rules First some terminology
A junction, also called a node or branch point, is is a point where three or more conductors meet
A loop is any closed conducting path

19. Kirchoff’s Rules Kirchoff’s Rules are basically two statements
1. The algebraic sum of the currents into any junction is zero
A sign convention:
A current heading towards a junction, is considered to be positive,
A current heading away from a junction, is considered to be negative
Be aware that all the junction equations for a circuit may not be independent of each other

20. Kirchoff’s Rules
2. The algebraic sum of the potential differences in any loop including those associated with emfs and those of resistive elements must equal zero
Procedures to apply this rule:
Pick a direction for the current in each branch
If you picked the wrong direction, the current will come out negative

21. Kirchoff’s Rules
Pick a direction for traversing a loop – this direction must be the same for all loops
Note that there is a third loop along the outside branches
As with the junction equations not all the loop equations will be independent of each other.

22. Kirchoff’s Rules
Starting at any point on the loop add the emfs and IR terms
An IR term is negative if we traverse it in the same sense as the current that is going through it, otherwise it is positive
An emf is considered to be positive if we go in the direction - to +, otherwise it is negative
Need to have as many independent equations as there are unknowns

23. Kirchoff’s Rules For loop I we have For loop II we have
Junction equation at a gives us
We now have three equations for the three unknown currents

24. Kirchoff’s Rules
Assume that the batteries are: e1 = 19 V; e2 = 6 V; e3 = 2 V
and the resistors are: R1 = 6W; R2 = 4W; R3 = 4W; R4 = 1W
you should end up with: I1 = 1.5 A; I2 = -0.5 A; I3 = 2.0 A
The minus sign on I2 indicates that the current is in fact in the opposite direction to that shown on the diagram
Complete details can be found here

25. RC Circuits
Up until now we have assumed that the emfs and resistances are constant in time so that all potentials, currents, and powers are constant in time
However, whenever we have a capacitor that is being charged or discharged this is not the case
Now consider a circuit that consists of a source of emf, a resistor and a capacitor but with an open switch
With the switch open the current in the circuit is zero and zero charge accumulates on the capacitor

26. RC Circuits Now close the switch
Initially the full potential will be across the resistor as the potential across the capacitor is zero since q is zero
Initially the full potential is across the resistor
The initial current in the circuit is then given by
As the current flows a charge will accumulate on the capacitor
At some time t, the current in the circuit will be I and the charge on the capacitor will be q

27. According to Kirchoff’s 2nd rule we have Using a counterclockwise loop
RC Circuits
According to Kirchoff’s 2nd rule we have
Using a counterclockwise loop
Solving for the current
As time increases, the charge on the capacitor increases, therefore the current in the circuit decreases
Current will flow until the capacitor has a charge on it given by

28. RC Circuits We remember that So we then have Rearranging we have
Setting up the integration we have
The resultant integration yields

29. RC Circuits
We exponentiate both sides of this last equation and rearrange to obtain
where Qf is the final charge on the capacitor given by Ce
The constant RC is known as the time constant of the circuit
We see that the charge on the capacitor increases exponentially

30. Example 3 I1 I3 I2 e R2 C R1
At t = 0 the switch is closed in the circuit shown. The initially uncharged capacitor then begins to charge.
What will be the voltage across the capacitor a long time after the switch is closed?
(a) VC = 0
(b) VC = e R2/(R1+ R2)
(c) VC = e
After a long time the capacitor is completely charged, so no current flows through it
The circuit is then equivalent to a battery with two resistors in series
The voltage across the capacitor equals the voltage across R2 (since C and R2 are in parallel)

31. RC Circuits The current in the circuit is given by and looks like
Note that is also how the voltage across the resistor behaves

32. RC Circuits – Charging Summary
For the simple RC circuit we have the following for the voltage drops across the capacitor and the resistor

33. RC Circuits
We now start from a situation where we have a charged capacitor in series with a resistor and an open switch
The capacitor will now act as a source of emf, but one whose value is not constant with time

34. RC Circuits We now close the switch and a current will flow
Kirchoff’s 2nd rule gives us
Rearranging we have
To find q as a function of time we integrate the above equation

35. RC Circuits
Exponentiation of both sides of the equation on the right yields
We see that the charge on the capacitor decreases exponentially

36. RC Circuits
The current in the circuit is obtained by taking the derivative of the charge equation
The quantity Q0 / C is just the initial voltage, Vo , across the capacitor
But then V0 / R is the initial current I0
So we then have that
The voltage across the resistor is given by

37. Example 4
The two circuits shown below contain identical fully charged capacitors at t = 0. Circuit 2 has twice as much resistance as circuit 1.
Compare the charge on the two capacitors a short time after t = 0
a) Q1 > Q2 b) Q1 = Q2 c) Q1 < Q2
Initially, the charges on the two capacitors are the same. But the two circuits have different time constants: t1 = RC and t2 = 2RC
Since t2 > t1 it takes circuit 2 longer to discharge its capacitor
Therefore, at any given time, the charge on capacitor 2 is larger than that on capacitor 1

38. Example 5 a) b) c) (a) t0 < ta (b) t0 = ta (c) t0 > ta R 3R C
The capacitor in the circuit shown is initially charged to Q = Q0. At t = 0 the switch is connected to position a.
At t = t0 the switch is immediately flipped from position a to position b.
a) Which of the following graphs best represents the time dependence of the charge on C? a) b) c)
b) Which of the following correctly relates the value of t0 to the time constant ta while the switch is at a?
(a) t0 < ta
(b) t0 = ta
(c) t0 > ta

39. Example 5 C a b R 3R
The capacitor in the circuit shown is initially charged to Q = Q0. At t = 0 the switch is connected to position a.
At t = t0 the switch is immediately flipped from position a to position b.
a) Which of the following graphs best represents the time dependence of the charge on C? a) b) c)
For 0 < t < t0, the capacitor is discharging with time constant t = RC
For t > t0, the capacitor is discharging with time constant t = 3RC,
i.e., much more slowly
Therefore, the answer is a)

40. Example 5 a) b) c) (a) t0 < ta (b) t0 = ta (c) t0 > ta R 3R C
The capacitor in the circuit shown is initially charged to Q = Q0. At t = 0 the switch is connected to position a.
At t = t0 the switch is immediately flipped from position a to position b. a) b) c)
b) Which of the following correctly relates the value of t0 to the time constant ta while the switch is at a?
(a) t0 < ta
(b) t0 = ta
(c) t0 > ta
We know that for t = ta, the value of the charge is e-1 = 0.37 of the value at t = 0
Since the curve shows Q(t0) ~ 0.6 Q0, t0 must be less than ta

41. Capacitors Circuits, Qualitative
Basic principle:
Capacitor resists rapid change in Q  resists rapid changes in V
Charging
It takes time to put the final charge on
Initially, the capacitor behaves like a wire (DV = 0, since Q = 0).
As current starts to flow, charge builds up on the capacitor
 it then becomes more difficult to add more charge
 the current decreases
After a long time, the capacitor behaves like an open switch.
Discharging
Initially, the capacitor behaves like a battery.
After a long time, the capacitor behaves like a wire.