Georg Simon Ohm was a German physicist known for his law describing the relationship between voltage, current, and resistance in electrical circuits. Through experiments connecting circuits to voltage sources and measuring voltage and current, Ohm determined that: 1) The higher the voltage at the source, the greater the resistance and voltage in the circuit. 2) If the strength of electrical current is less, then the electrical resistance is higher. Ohm eventually concluded that electrical sources themselves produce some resistance due to the wires or electrolytes within them, known as internal resistance. His full law, known as Ohm's Law for closed circuits, states that the strength of the current in a closed circuit is equal to the quotient of the electrom

1. Georg Simon Ohm

2. Repeat the previous lesson
1. What is electrical resistance?
Answer: This is an event aimed to counteract the movement of charged
particles
2. How to define Ohm's law for part circuit?
Answer: The strength of an electric current to a part of the electric circuit is
proportional to the quotient of voltage and inversely proportional to the
resistance to electrical current portion of the circle.
3. How can I tie the resistors in the circuit?
Answer: The serial and parallel connection.

4. - Ohm“s law for closed circuit -
In order, we explained how to Ohm came to the conclusion that the source exists
electrical resistance, we need to perform experiments:
1. Let's put the circuit as in the diagram!
-
We will increase the voltage of electricity at the source.
-
What we see and what we conclude?
This is Ohm's law for part of the circuit!
For Ohm“s law, the following relations:
Conclusion 1: The higher the voltage at the source,
the greater the resistance and voltage in the circuit.
Conclusion 2: The amount of electricity is higher, the higher the voltage,
If the strength of the electrical current is less, then the electrical resistance higher.

5. Ohm“s law for closed circuit
• Ohm has long worked on
the part of an electrical
circuit, until he saw that
the electrical source that
gives power has some
resistance, because he
made ​of wire or
electrolyte that also
produce resistance.
• Later voltimetar set to an
electrical source and
began the examination:

6. 2. Let's put the circuit as in the diagram!
- We know that the EMS (electromotive force)
the potential difference at the poles sources.
We measure it in volts and represents the
total energy per unit charge that can be used
from the source.
- When we connected the circuit voltimetar we
will read a certain voltage.
- Voltimetar now measure the voltage source
at the poles, where the circuit, and the
current source starts. This voltage is always
lower than EMS batteries because charge
carriers encounter resistance moving the
terminals inside the battery.
- This resistance we give the name of the
internal resistance of the battery (source)
- Marks it with a lowercase "r" unit (Ω)

7. - From this we can conclude that the real source of resistance to voltage drop of
electricity.
We can mathematically represented as follows:
•
•
Uv
I R
Uu
I r
E
I (R r)
•
•
•
•
- Voltage drop on the outside of the
elecrical circuit;
- voltage drop across the inner part
of the electrical circuit;
R-external resistance (resistance
bulbs);
I-strength electrical current;
E-EMS (elktromotive force);
r-internal resistance sources
This term represents Ohm's law for closed circuit.
Ohm's law for closed circuit as follows:
The strength of the current in a closed circuit is equal to the quotient of the
electromotive force and the total resistance.

8. Now I will show you how electrical
source produces resistance

9. Example No.1.
1. On the electrical source electromotive force 4V and internal resistance 0.2 Ω
resistor is connected whose resistance 5Ω. Identify:
a) Find the strength of the electrical current through the circuit "I";
b) Time "t", which will pass through a round amount of electricity 20C.
E=4V
r=0,2Ω
R= 5Ω
q=20C
I=? ; t=?
E
4V
a) I
0,77 A
R r 5 0,2
q
q 20C
b) I
t
25,97 s
t
I 0,77 A

10. Example No. 2.
Power source 12V electromotive force and internal resistance of 0.8 Ω is connected to
a circuit as shown in the diagram. What is the strength of the current in the circuit?
Solution:
E=12V; r=0,8 Ω; R1=10Ω; R2=6 Ω
R=R1+R2=6+10=16Ω
6Ω

11. Repeat the whole lesson
1. Will the electrical source produces electrical
resistance in the circuit
Answer: Yes, the source generates electric resistance.
2. As we mark the internal resistance of the power
source?
Answer: The internal resistance sources denoted with "r„.
3. What is the definition of Ohm's law for closed circuit?
Answer: The strength of the current in a closed circuit is
equal to the quotient of the electromotive force and the
total resistance.

12. Homework
• Physics Book 9, page 43 and 44: Check the
knowledge! Answer! Calculate! Tasks 1 to 8