DC Circuits

 Resistors in series are connected end to end
(one path thru them and back to the V)
 Any charge passing thru R
R
2
1 also passes thru and so on… In series the same current, I, exists in every part of the circuit.
(Conservation)
Equivalent resistance, R eq is re drawing the circuit with ONE resistor in place of all the others. For a series circuit,
R eq
R eq
= R
1
+ R
2
+ R
3
+… is the sum of the resistors in the circuit.

 Potential difference at the battery, V, is established across each resistor in a series circuit.
 Since the current in a series circuit is the same everywhere , the Voltage dropped across each resistor is found by Ohm’s law:
V = I * R


V
1
= IR
1
…)= V batt and V
2
= V
1
= IR
+ V
2
2 so V
1+2+…
+ V
3
V batt
= I*R eq
= I (R
1
+ R
2
+

 When resistors are connected in parallel, the current from the source splits into separate branches.
 What happens if you disconnect one device in a parallel circuit?
 The total current I that leaves the battery is the sum of the current in each branch.
 I
T
= I
1
+ I
2
+ I
3
+ …

 When resistors are connected in parallel, each is supplied the same voltage (have the same potential as the battery).
 I
1
=V/R
1 and I
2
=V/R
2 and I
3
=V/R
3
…
 The equivalent resistance must draw the same current as shown by I=V/R eq
 The equivalent resistance is found by:
1
R eq
1
R
1
1
R
2
1
R
3
...

 An equivalent resistance is like combining all resistors and re drawing the circuit with the one equivalent resistor. (I, V, and P remain unaffected).
 If only 2 resistors are parallel, the R eq product / sum (R
1
R
2
) / (R
1
+R
2
)
= the


If the two resistors are the same value, the
R eq is HALF of one of them.
3Ω and 3Ω parallel makes a R eq of 1.5Ω.

 Draw a circuit with a voltage of 12.0V where two resistors valued at 500Ω and
700 are in parallel with each other and in series with a third resistor valued at
400Ω.
 What is the current flowing through the
500Ω resistor?

 To find out all information about a parallel circuit, we do things in a different order than with series circuits.
 First we know the voltage at the battery is the same on every branch.

Next we can use Ohm’s law to find the current on each branch.
 Sum the branch currents to get battery current.
 Find R eq
.

 Electromotive force (not really a force) is a source (like a battery or electric generator) that converts one type of energy (chemical, mechanical, light, etc…) into electric energy.
 The potential difference between the terminals of the source when NO CURRENT flows through it is called the emf. Symbol E.
 When current is drawn from a battery, the voltage drops. A battery has some internal resistance , r.


 The terminal voltage is what is measured between the terminals of the battery when it is not connected to the circuit. At this time, the voltage between points a and b (the terminals) is = to the emf. (determined by the chemical reactions in the battery.)
 V ab
= E
 When current flows from the battery, there is a drop in voltage = to I r . So V ab
=E – I r .

 Sometimes circuits get very complicated and require more than Ohm’s Law to analyze.
 G.R. Kirchhoff (1824-1887) developed two rules based on the laws of conservation of charge and energy.

Kirchhoff’s first or junction law states: at any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction . (What goes in must come out!)


Kirchhoff’s second or loop rule is based on the conservation of energy.

“ The sum of the changes in potential around any closed path of a circuit must be zero .”
I = V / R =
12.0V / 690 Ω
= 0.0174A
a
400Ω b
290Ω c e
12.0V
d

 Point. e, the positive side of the battery is at a high potential and point d, the negative side is at a lower potential.
 We follow the current around the circuit
(positive test charge?) and drop off potential
(voltage) along the way like the graph on p 565 shows.
 The decrease in voltage between the two ends of a resistor is called a voltage drop and gets a negative sign using Kirchhoff’s loop rule:
 V ba
= V b
– V a
= -6.96 v

 Changes in potential around the circuit are plotted below.
12v
6.96v
V
5.04v
0v e a b
400 Ω 290Ω 12.0v
12.0v
c d e

 Please do ch 19 Rev p 591 #s 1, 3, 5, 7,
& 9.