Important Current and Electricity Formulas for JEE

The flow of charges through a wire is known as an electric current. Amperes are the units of measurement for current. The circuit must be closed in order for the charge to pass into it. To put it another way, the journey from the power source to the circuit and back must be uninterrupted. If there is a potential gap, current flows through the circuit. Any charge movement from one region to another is considered current.

When two bodies with different potentials are connected by a wire, free electrons flow from one to the other until both bodies reach the same potential, at which point the current ceases to flow. Current flows until a potential difference is evident across a conductor. Voltage is the difference in electric potential and is measured in volts.

Resistance is the force that opposes the passage of electric current and is measured in ohms. The heat energy can be converted from the electric current lost owing to resistance. The heat generated is utilised to power an electric stove.

The List of Important Current Electricity Formulas for Jee Mains

The list of important current electricity formulas for JEE mains and current electricity important formulas for NEET is as follows:

1. Current

I = n x e x A x V\[_{d}\]

n: number of free electrons per unit volume

A: area of cross-section

e: charge of the electron

Vd: drift velocity

2. Current Density

j = \[\frac{i}{A}\] = σ x E

i: current

A: cross-sectional area

E: electric density

σ: conductivity

3. Resistance of the Wire

R = \[\frac{\rho \times I}{A}\]

Where \[\rho\] =  \[\frac{1}{\sigma}\]  

ρ is the resistivity

σ is the conductivity

4. Temperature Dependence of Resistance

R = R\[_{0}\] (1 + \[\alpha\] x ΔT)

R\[_{0}\]: original resistance

R: resistance at temperature T

α: temperature coefficient of resistance

5. Ohm’s Law

V = IR

V: voltage

I: current

R: resistance

6. Kirchhoff’s Law

(i) Kirchhoff’s First law 

\[\sum\] i = 0 at any junction.

(ii) Kirchhoff’s Second Law

\[\sum\] iR = 0 in a closed circuit.

7. Resistance 

R = \[\frac{V}{i}\] 

i: current

V: potential Difference

8. Conductance

G = \[\frac{1}{R}\]

R: Resistance 

9. Electric Power

P = \[\frac{V^{2}}{R}\] = I\[^{2}\]R 

10. Galvanometer as an Ammeter

G = \[\frac{(i-i_{g})S}{i_{g}}\]

G: resistance of the galvanometer

S: shunt resistance

i: maximum current measured by the ammeter

i\[_{g}\]: maximum current passed through the galvanometer

11. Galvanometer as a Voltmeter

V = i\[_{g}\](R + G)

G: resistance of the galvanometer

i\[_{g}\]: maximum deflection of the galvanometer

R: high resistance connected in series

12. Time Constant in RC Circuit

τ = RC

13. Resistance in Parallel

1Req = 1R1 +  1R2\[\frac{1}{R_{eq}}\] = \[\frac{1}{R_{1}}\] + \[\frac{1}{R_{2}}\]

14. Resistance in Series

R\[_{eq}\] = R\[_{1}\] + R\[_{2}\]

Solved Examples on Current Electricity Formulas for JEE Mains and Current Electricity Important Formulas for NEET

Ex.1. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 5% each, then the error in the value of resistance of the wire is,

i. 6% 

ii. 5%

iii. 10%

iv. 4%

Solution:

R = \[\frac{V}{I}\] and \[\Delta\]R = \[\frac{\Delta V}{\Delta I}\]

Now,

\[\frac{\Delta R}{R}\] = ±(\[\frac{\Delta V}{V}\] + \[\frac{\Delta I}{I}\])

= ± (5 + 5)%

= ± 10%

Hence the value of resistance of a wire is 10%.

Ex.2. The current in the primary circuit of a potentiometer is 0.5A. The specific resistance and cross-section of the potentiometer wire are 4 x 10\[^{-7}\] ohm meter and 8 x \[^{-7}\] m\[^{2}\] respectively. The potential gradient will be equal to,

i. 0.2 V/m 

ii. 0.25 V/m

iii. 0.5 V/m

iv. 0.15 V/m

Solution: 

Given i = 0.5A, ρ = 4 x \[^{-7}\] Ωm, A = 8 x 10\[^{-7}\]m\[^{2}\] 

x = \[\frac{i \rho}{A}\] = \[\frac{0.5 \times 4 \times 10^{-7}}{8 \times 10^{-7}}\]

x = 0.25 V/m

Hence the potential gradient is 0.25 V/m.

Ex.3. Drift speed of electrons, when 2.5 A of current flows in a copper wire of cross-sectional area 5 mm\[^{2}\] is v. If the electron density of copper is 8 x 10\[^{28}\]/m\[^{3}\] the value of v in mm/s is close to (Take charge of an electron to be e = 1.6 x 10\[^{-19}\] C)

(a) 0.4

(b) 0.02

(c) 3

(d) 0.04

Solution:

I = 2.5

n = 8 x 10\[^{28}\]/m\[^{3}\]

e = 1.6 x 10\[^{-19}\]

A = 2.5 x 10\[^{-6}\]

I = n x e x A x V\[_{d}\]

V\[_{d}\] = \[\frac{I}{n \times e \times A}\] = \[\frac{2.5}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-6}}\]

V\[_{d}\] = \[\frac{2.5}{64 \times 10^{3}}\]

V\[_{d}\] = 0.039 x 10\[^{-3}\] m/s

V\[_{d}\] = 0.04 mm/s