Chapter 19
The Giancoli onLine Tutor

DC Circuits

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I. Circuits with Resistance

Series and Parallel Circuits

At the right two simple circuits are shown. The first is a series circuit and the 2 resistors can be replaced by a single resistor whose value,R_eq, is given by:

R_eq= R₁+R₂

If R₁= 10 W and R₂= 20 W, then R_eq= 30 W. The second circuit is a parallel circuit and the two resistors can be replaced by a single resistor whose value, R_eq, is given by

1	=	1	+	1
R_eq	=	R₁	+	R₂

If R₁= 10 W and R₂= 20 W, then R_eq= 6.67 W.

Series-Parallel Circuits

At the left is a combination of series and parallel. First find the equivalent resistance for the parallel resistors and then find the equivalent series resistance.
If R₂= 10 W and R₃= 20 W, then R_eq= 6.67 W. If R₁= 5 W and R_eq= 6.67 W, then R^'_eq= 11.67 W.

Interactive Activity
Sometimes the circuit can be drawn in a confusing manner. Consider the circuit on the right.

Any resistors that are connected by a wire with no junction between them are series resistors and should be reduced accordingly.

In an ideal circuit all wires have zero resistance. Any resistors that have two junctions in common with nothing inbetween are actually in parallel.

And can be reduced to a single value.

Kirchhoff's Rules

Some circuits can't be analyzed in the above manner.

For example, if you have sources of emf in different branches, as in the circuit on the left, the above methods would not work. Here you have to use.
1) The sum of all currents going into a junction must equal the sum leaving the junction.

The first step is shown on the left. Pick a junction and draw in currents in each separate branch (label them 1, 2, 3,...). Indicate the directions of the current by drawing arrows. (Your choice here is not important, if you 'guess' the wrong direction the current will come out negative). Thus we find

(1) I₁+I₂= I₃

2) The sum of potential changes around any closed loop in the circuit must add to zero.

The figure shows three possible loops. How many are needed? There are three unknowns, I₁, I₂ and I₃, so we need three equations; we already have 1, so we need two more. We can use two of the three loops shown. To be specific choose "1" and "2". Think of a current loop as a ski resort. Resistors are Slopes, Currents are Skiers, and batteries Ski Lifts.

When you go through a resistor in the direction of the current you are going downhill. (The potential change is negative).
When you go through a resistor against the direction of the current you are going uphill. (The potential change is positive).
When you go through a battery from '+' to "-" you are riding the ski lift down. (The potential change is negative).
When you go through a battery from '-' to "+" you are riding the ski lift up. (The potential change is positive).

Interactive Example 2

Starting in the upper right hand corner of loop 1 we have

(2) 10 - 5I₁ -20I₃ + 3 = 0
OR

(2) =

Now complete the second loop starting in the upper right hand corner and proceeding counter-clockwise . If you enter an incorect value it will self distruct when you click on the next entry allowing you to correct your mistake.

(3) =

To solve for the currents:

Add the second and third equations above. Do it! click on the field below to see if your result was correct.

Substitute into this equation I₂= I₃-I₁ to obtain a new equation. Do it! Check your result below.

To eliminate I₃ and thus find I₁ multiply this last equation by 2 and add it to equation 2. Try it and check your result below.

Substitute this into equation (2) to find I3. Substitute I3 into equation 3 to find I2. Finally as a check, substitute all three into Equation 1 to verify your results. After you have done this click below.

Real Batteries

A real battery in addition the being a seat of emf, E, also has internal resistance, r. When no current is drawn from the battery the terminal voltage, V, would be equal to E, however, when current, I, is drawn from the battery, the terminal voltage drops to E-Ir.

Interactive Example
What is the terminal voltage, V, on the battery in the upper leg of the circuit on the left. Need a hint.

II. Capacitor Circuits

Series and Parallel Capacitor Circuits

At the right two simple capacitor circuits are shown. The first is a series circuit and the 2 capacitors can be replaced by a single capacitor whose value,C_eq, is given by:

1	=	1	+	1
C_eq	=	C₁	+	C₂

If C₁= 10 mF and C₂= 20 mF, then C_eq= 6.67 mF.

The second circuit is a parallel circuit and the two capacitors can be replaced by a single capacitor whose value, C_eq, is given by

C_eq= C₁+C₂

If C₁= 10 mF and C₂= 20 mF, then R_eq= 30 mF.

Series-Parallel Capacitor Circuits

Interactive Example 3
In the figure :
C₁= 5 mF, C₂= 10 mF and C₃= 20 mF.
This is a combination of series and parallel capacitors. First find the equivalent resistance for the parallel resistors and enter the results below. Click outside and if you get it right the picture will change, if not you will get a hint. and then find the equivalent series resistance .

III. Resistor - Capacitor Circuits

Two RC circuits are shown. In circuit A the capacitor is uncharged and its voltage is correspondingly zero. When the switch is closed the capacitor will 'gradually' charge and its voltage will eventually rise to E.

Interactive Example 2
Close the switch below and everytime you press advance the time will increase by 0.1 s showing the volatage on the capacitor at that time. If the capacitor is 6 mF and E=10 V what is the resistance?

t= V=

Need a hint.

In circuit B The Capacitor is charged and has a voltage of 20 volts when the switch is closed current will flow and the capacitor's voltage will gradually drop to zero. If the Resistance is 400000W what is the capacitance?

Close the switch below and everytime you press advance the time will increase by 0.1 s showing the volatage on the capacitor at that time.

t= V=

Need a hint.