ELECTRICITY PART 2

ELECTRICITY PART 2

CBSE class x science chapter 12 ELECTRICITY NOTES
Electric Field(E): It is the space around an electric charge in which its effect can be observed by another charge particle.
Mathematical : E=F/q
Electric Field
Representation of Electric Field by Source and Test Charge

S I unit of electric field is NC-1
CBSE class x science chapter 12 ELECTRICITY NOTES
Electric Potential: The amount of work done in bring a unit positive charge from infinity to a point inside electric field of source charge without acceleration. 

Diagram:
Electric Potential
Electric Potential

Mathematically: If external force F applied on test charge q and it displace to a distance d, 

then work done W = F x d

According to definition V=W/q  


S I unit of electric potential is volt.
                                1V=1J/1C

Definition of 1 volt: If one joule of work is done in bring 1 coulomb of charge from infinity to a point inside electric field without acceleration.

CBSE class x science chapter 12 ELECTRICITY NOTES

Electric Potential Difference: The amount of work done in bring a unit positive charge from one point to another in electric field of source charge without acceleration.

Diagram:
Electric Potential Difference
Electric Potential Difference
Mathematically:

 The charge q move from point A to point B inside electric field ΔV= VB-VA=WAB/q
CBSE class x science chapter 12 ELECTRICITY NOTES
S I unit of electric potential difference is volt (1V=1J/1C).

Definition of 1 volt: If one joule of work is done in bring 1 coulomb of charge from one point to another with in electric field with out acceleration.

Device used to measure Potential difference is Voltmeter.
Voltmeter is always joined in parallel with circuit.
Ideal voltmeter has infinite resistance.
CBSE class x science chapter 12 ELECTRICITY NOTES
OHM's LAW: If physical condition remains unchanged,the potential difference applied across a conductor is directly proportional to the current following through the conductor.
i e   V  ∝ I
V = R I, Where R is constant of proportionality or Resistance.

 R = V/I

Resistance is numerically equal to the ratio of potential difference applied across the conductor to the current following in the conductor.

ACTIVITY:

AIM: To verify Ohm's Law experimentally.

APPARATUS: Voltmeter , ammeter , connection wire, battery , key , conductor(resistance) , rheostat. Or (Ohm's Law apparatus).

THEORY:If physical condition remains unchanged,the potential difference applied across a conductor is directly proportional to the current following through the conductor.
i e   V  ∝ I

FORMULA USED: V = I R
              R = V / I

CIRCUIT DIAGRAM:
OHM'S LAW CIRCUIT DIAGRAM
Circuit Diagram of OHM'S LAW
PROCEDURE:
CBSE class x science chapter 12 ELECTRICITY NOTES
  1. Join the connections as shown in circuit diagram.
  2. Check the connection by inserting key and voltmeter and ammeter will give deflection.
  3. Find the least count of voltmeter and ammeter.
  4. Now changing the value of voltage either from battery or from rheostat,note the corresponding value of current.
  5. Repeat step 4 for at least five set of reading.
OBSERVATIONS:
S N
POTENTIAL DIFFERENCE V (V)
CURRENT I (A)
RESISTANCE  R=V/I (Ω)
1
2
4
0.5
2
4
8
0.5
3
6
12
0.5
4
8
16
0.5
5
10
20
0.5

GRAPH:
Graph between V and I
CONCLUSION:
From the table the the value of V and I are proportional and the ratio of V/I is also constant i e R is constant.
The graph between V and I is also straight line so it proves that V  ∝ I.
CBSE class x science chapter 12 ELECTRICITY NOTES
Graph of Ohm's Law and SLOPE:
I and V graph for Ohm law
Graph between (V vs I) and (I vs V).
SLOPE FOR V vs I
SLOPE FOR I vs V
SLOPE= tan θ = ΔV/ΔI=R
SLOPE= tan θ = ΔI/ΔV=1/R

QUESTION/NUMERICAL BASED ON TOPICS:

Q How much energy is given to each coulomb of charge passing through a 6 V battery?

 Ans Potential Difference (V) = 6 V
 Charge (q) = 1 C
Energy = total work done = W = q x V= 1 x 6 = 6 V [because V=W/q ]

 Q Calculate the resistance of an electric bulb which allows a 10 A current when connected to a 220 V power source ?

 Ans Given , I = 10 A , V= 220 V

 R = V/I = 220/10 = 22 Ohm. 


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