1. If a net charge Q, flows across any cross-section of a conductor in time t, then the current I, through the cross-section is
I = Q/t
Q = It
2. The electric potential difference between two points in an electric circuit carrying some current is the work done to move a unit charge from one point to the other –
Potential difference (V) between two points = Work done/Charge
V = W/Q
W = VQ
3. Q = n x Charge on 1 electron
When a steady current flows through a conductor, the electrons in it move with a certain average ‘drift speed’.
4. If the current I, flowing in a metallic wire and the potential difference across its terminals is V .
Then potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. This is called Ohm’s law.
V a I
Þ V = RI or, I = V/R
5. Resistance of the conductor depends (i) on its length, (ii) on its area of cross-section, (iii) on the nature of its material and (iv) temperature
R a l/A
R = r l/A Or, r = RA/l
6. If resistors joined in series:
V = V1 + V2 + V3
but I = I1 = I 2 = I 3
Then R = R 1 + R 2 + R 3
6. If resistors joined in Parallel:
V = V1 = V2 = V3
but I = I1 + I 2 + I 3
Then 1/R = 1/R 1 + 1/R 2 + 1/R 3
7. If a current I flowing through a resistor of resistance R. and the potential difference across is V for time t sec
Then, the work done in moving the charge Q through a potential difference V is
W= VQ.
But, Q = It
W = V I t ------(i)
Now, Power = work done / Time
P = W/t
{or, W = Pt [The energy supplied to the circuit by the source in time t is P × t = V I t]
P = V I t /t [ Using eq. (i) ]
P = VI ----------- (ii)
The amount of heat produced in time t = H
H = the energy supplied to the circuit by the source in time t = V I t
Applying Ohm’s law, V = IR
H = I2 R t
Note: heat produced in a resistor is
(i) Directly proportional to the square of current for a given resistance,
(ii) Directly proportional to resistance for a given current, and
(iii) Directly proportional to the time for which the current flows through the resistor.
8. Electric Power: The rate at which electric energy is dissipated or consumed in an electric circuit is called electric power.
The power P is given by P = VI Using, V = IR
Þ P = I2R = V2/R
Also using, V/R = I
Þ P = V2/R
9. The commercial unit of electric energy is kilowatt hour (kW h) = 1 unit.
10. 1 kW h = 1000 watt × 3600 second = 3.6 × 106 watt second = 3.6 × 106 joule (J)
Now start practicing these Numerical: Click here To View
I = Q/t
Q = It
2. The electric potential difference between two points in an electric circuit carrying some current is the work done to move a unit charge from one point to the other –
Potential difference (V) between two points = Work done/Charge
V = W/Q
W = VQ
3. Q = n x Charge on 1 electron
When a steady current flows through a conductor, the electrons in it move with a certain average ‘drift speed’.
4. If the current I, flowing in a metallic wire and the potential difference across its terminals is V .
Then potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. This is called Ohm’s law.
V a I
Þ V = RI or, I = V/R
5. Resistance of the conductor depends (i) on its length, (ii) on its area of cross-section, (iii) on the nature of its material and (iv) temperature
R a l/A
R = r l/A Or, r = RA/l
6. If resistors joined in series:
V = V1 + V2 + V3
but I = I1 = I 2 = I 3
Then R = R 1 + R 2 + R 3
6. If resistors joined in Parallel:
V = V1 = V2 = V3
but I = I1 + I 2 + I 3
Then 1/R = 1/R 1 + 1/R 2 + 1/R 3
7. If a current I flowing through a resistor of resistance R. and the potential difference across is V for time t sec
Then, the work done in moving the charge Q through a potential difference V is
W= VQ.
But, Q = It
W = V I t ------(i)
Now, Power = work done / Time
P = W/t
{or, W = Pt [The energy supplied to the circuit by the source in time t is P × t = V I t]
P = V I t /t [ Using eq. (i) ]
P = VI ----------- (ii)
The amount of heat produced in time t = H
H = the energy supplied to the circuit by the source in time t = V I t
Applying Ohm’s law, V = IR
H = I2 R t
Note: heat produced in a resistor is
(i) Directly proportional to the square of current for a given resistance,
(ii) Directly proportional to resistance for a given current, and
(iii) Directly proportional to the time for which the current flows through the resistor.
8. Electric Power: The rate at which electric energy is dissipated or consumed in an electric circuit is called electric power.
The power P is given by P = VI Using, V = IR
Þ P = I2R = V2/R
Also using, V/R = I
Þ P = V2/R
9. The commercial unit of electric energy is kilowatt hour (kW h) = 1 unit.
10. 1 kW h = 1000 watt × 3600 second = 3.6 × 106 watt second = 3.6 × 106 joule (J)
Now start practicing these Numerical: Click here To View
| 10th Electricity – Remember these terms before solving Numerical problems |
