Current electricity (revision sheet)

1. Ohm’s Law : V=IR


2. Resistivity : ρ=(A/l)R

3. Temperature coefficient of resistance : α=( ρ- ρ₀)/t = (R-R₀)/t

4. Resistance in series : R=R₁+R₂+R₃+…….

5. Resistance in parallel : R = 1/R₁+1/R₂+……

6. Terminal potential difference V=IR=E-Ir

7. Cells in series E=E₁+E₂+…..

8. Electrical energy W=EIt

9. Heat produced in a resistor H=I2Rt = VIt

10. Power dissipated in resistor P=VI + I₂R

11. Maximum current in n rows of m cells each : I_max=mE/2R = nE/2r

12. Wheatstone bridge when balanced : P/Q = R/S

13. Kirchoff’s Law : (1) For junction I₁+I₂+I₃+……=0 (2) For a loop I₁R₁+I₂R₂+I₃R₃+……= sum of potential drops

14. Shunt resistance required to convert galvanometer into ammeter to read upto I amperes: S=(I_gG)/(I-I_g) where G is the galvanometer resistance and Ig is the current through galvanometer to produce full scale deflection

15. Current when n cells are connected in series I=nE/(R+nr)

16. Current when n cells are connected in parallel I=nE/(nR+r)

17. For balanced potentiometer E₁/E₂=l₁/l₂

How to simplify circuits with resistors

1. In any given circuit first of all recognize the resistances connected in series then by summing the individual resistances draw a new, simplified circuit diagram.

For series combination of resistances equivalent resistance is given by the equation
R_eq= R₁ + R₂+R₃
The current in each resistor is the same when connected in parallel combination.

2. Then recognize the resistances connected in parallel and find the equivalent resistances of parallel combinations by summing the reciprocals of the resistances and then taking the reciprocal of the result. Draw the new, simplified circuit diagram.
(1/R)=(1/R₁)+(1/R₂)+(1/R₂)

Remember that for resistors connected in parallel combination ‘The potential difference across each resistor is the same’.

3. Repeat the first two steps as required, until no further combinations can be made using resistances. If there is only a single battery in the circuit, this will usually result in a single equivalent resistor in series with the battery.

4. Use Ohm’s Law, V= IR, to determine the current in the equivalent resistor. Then work backwards through the diagrams, applying the useful facts listed in step 1 or step 2 to find the currents in the other resistors. (In more complex circuits, Kirchhoff’s rules can be applied).

Color code of carbon resistance

• Commercially resistors of different type and values are available in the market but in electronic circuits carbon resistors are more frequently used
• In carbon resistors value of resistance is indicated by four colored bands marked on its surface as shown below in figure

• The first three bands a,b.c determine the value of the resistance and fourth band d gives the tolerance of the resistance
• The color of the first and second band respectively gives the first and second significant figure of the resistance and third band c gives the power of the ten by which two significant digits are multiplied for obtaining the value of the resistance
• value of different colors for making bands in carbon resistors are given below in the table

Color Figure(first and second band) Multiplier(for third band) tolerance
• For example in a given resistor let first strip be brown ,second strip be red and third be orange and fourth be gold then resistance of the resistor would be 12 x 10³ (± 5%

For more notes in physics visit physicscatalyst.com

Electric resistance and Resistivity

(A) Resistivity

In the previous post we derived that current density is
j = nqv_d
where v_d is the drift velocity.

Current density in general depend on electric field and for metals current density is nearly proportional to the electric field. (Results can be derived using theory of metallic conduction.)

Thus for metals ratio of E and j is constant and for a particular material its resistivity ρ is defined as the ratio of magnitude of electric field to current density,
ρ = E/j
This relationship is known as Ohm's law discovered by german physicist Georg Simon Ohm (1787-1854) in 1826.

Greater would be the resistivity of a given material greater field would be required to establish a given current density in the material or we can say that smaller would be the current density for a given field.

Unit of resistivity is Ωm (ohm. meter).

Materials having zero resistivity are known as perfect conductors and those having infinite resistivities are known as perfect insulators. Real materials lie between these two limits.

Metals and alloys are materials having lowest resistivities and are good conductors of electricity.

Insulators have resistivities many times (of the order of 10²²) greater then that of metals.

Reciprocal of resistivity is conductivity. Unit of conductivity is (Ωm)^-1.

Metals or good conductors of electricity have conductivity greater than that of insulators.

Semiconductors are those materials which have resistivities intermediate between those of metals and insulators.


(B) Resistivity and temperature

Resistivity of a conductor depends on a number of factors and temperature of the metal is one such factor. As the temperature of the conductor is increased its resistivity also increases.

For small variations in temperature resistivity of materials is given by the relation
ρ(T) = ρ(T₀)[ 1 +α(T-T₀)]
where, ρ(T) and ρ(T₀) are resistivities at temperature T and T₀ respectively and α is constant for a given material which also depends on temperature to a small extent. This constant α is known as temperature coefficent of resistivity.

(C) Resistance

We already know that for a conducror relation between electric field E and current density is given as
E = ρj
where ρ is a constant independent of E.

When we study electric circuits we are more interested in the total current in a conductor rather then current density j and more interested in knowing the potential difference between the ends of the conductor than in Electric field becaude current and potential difference are easier to measure then j and E.

Consider a conducting wire of length l and uniform crossectional area A. If V is the potential difference between both the ends of the wire then electric field inside the conductor would be
E = V/l
If i is the current flowing inside the wire then current density is given by
j = i/A
putting these values in Ohm's law ρ = E/j we get
V = ρi (l/A)
or , V=Ri
where, R=ρ(l/A)
which is known as resistance of a given conductor.

Unit of resistance is ohm or volt per ampere.

Thus how much current will flow in a wire not only depends on the potential difference between two ends of the wire but also on the resistance offered by the conductor to the flow of electric charge.

From the above discussion we can easily conclude that The resistance of a wire depends both on the thickness and length of the wire and also on its resistivity.

Thick wires have less resistance then thin ones and longer wires have more resistance then shorter ones.

Since the resistivity of a marerial varies with temperature, the resistance of any particular conductor also varies with temperature. For temperature ranges that are not too great, this variation is approximately a linear relationship, analogous to the one we learned for resistivity
R(T) = R(T₀)[1 + α(T - T₀)]
In this equation. R (T) is the resistance at temperature T and R(T₀) is the resistance at temperature T₀. The temperature coefficient of resistance α is the same constant that appears in case of resistivity.

In the next post we'll do some worked examples related to this topic

Electric Current and Current Density

- Electric current is the motion of electrically charged particles from one region to another. This motion of electric charges takes place within electric circuit which is a conducting path forming a closed loop.
- In a conductor electric charge will flow from its one end to another if and only if both the ends of the conductor are at different electric potentials.
- Continous flow of electric current in a conductor for a relatively long period of time can be attained using battries which could supply continous flow of charge at low potentials.

Electric current
- Here in this section we would discuss about the electric current in conductors.
- When electric field inside the conductor is zero then there would be no net flow of current in the conductor because in this case electrons in the conductors moves about randomly leaving no net flow of charge in any direction and without the flow of charge there would be no net electric current.
-To maintain a constant current in a conductor we would have to ensure that a constant and steady electric field is established inside the conductor in order to maintain a force on the mobile charges in the conductor.
-Once the electric field is maintained inside the conductor charged particles in the conductors are now under the influence of driving force F = qE.
- In an conductor charged particles undergoes frequent inelastic collision with fixed massive ions in the conductor and undergoes random change in the direction of motion.
- Hence on an average charged particle moves in the direction of driving force with an average velocity known ad drift velocity.
- Electric current is defined as the quantity of charge ΔQ flowing through cross-sectional area A in time interval Δt . Thus,
I_av = ΔQ/Δt
which is the average current flow in time Δt.
- If dQ is the amount of charge flowing in infinitesimally time interval dt through a cross-sectional area of the conductor then instantaneous current I is defined as
I = dQ/dt
- Electric current is a scalar quantity.
- SI unit of current is Ampere (A) where 1A is one coulomb per second.

Drift velocity and Current density - Consider a portion of a conductor of cross-sectional area A. Also consider a small section of conductor of length Δx. Now volume of conductor under consideration is AΔx.
- If there are n number of mobile charge carriers per unit volume then total charge in the section under consideration is
ΔQ = (number of charge carriers in the section) x (charge per carrier)
= (nAΔx)q .................1
where q is the amout of charge on each carrier.
- If charge carriers are moving with speed v_d, then distance travelled by the charge carriers is Δx = v_dΔt. Putting this in equation 1 we have,
ΔQ = (nAv_dΔt)q
- Now current in the conductor is
I = ΔQ/Δt
thus,
I = nqv_dA ....................2
where v_d is average velocity known as drift velocity as defined earlier.
- Current density j is defined as current per unit cross-sectional area. Thus from 2
j = I/A = nqv_d
Current is a scalar quantity but current density can also be defined too include both magnitude and direction. Thus vector current density is
j = nqv_d
Direction of current density is same as the direction of electric field.
- Unit of current density is A.m^-2.
- Current density tells us about how charges flow at a certain point and also about the direction of the flow at that point but current describes how charges flow throughout an extended object.

Solved example
In this solved example we will lern to apply the principle of current to the problems.

Question:-Amount of charge that passes through a certain conducting wire in 4 sec is 6.5 C. Find (a) the current in the wire and (b) number of electrons that passes through the wire in 8 sec.
Solution : -
(a)
Problem solving strategy
1. From the lesson learned about electric current identify the equation for calculating current when charge and time are given.
2. Put the values of charge and time at respective place and calculate the answer.

In this case Q = 6.5 C and t = 4 sec.
now I = q/t = 6.5C/4 sec = 1.625 A
which is the required answer.

(b)
Problem solving strategy
1. Here we have to find total charge through the wire in a given time.
2. Total charge through the wire would be equal to number of electrons through the circuit multiplied by the charge on each electron.

Here we have to find the number of electrons passing through the wire in 8 sec. If q_e is the amount of charge on a single electron and total n number of electrons passes through the wire then
I = nq_e/t
or, N = It/q_e
putting values of I , t and q_e = 1.6 x 10^-19 and calculating we get
N = 8.125 x 10¹⁹

Same way problems related to current density and drift velocity can be solved.