Electric Current IIT JEE Study Material

The time rate (t) of flow of charges (q) through any cross-section is called an electric current. It is a scalar quantity and its SI unit is ampere (A). If charges \(\mathbf{\Delta \;Q}\) passes through a cross-section in time \(\mathbf{\Delta \;t}\) then,

Electric Current \(\mathbf{i_{av}\;=\;\frac{\Delta \;Q}{\Delta \;t}}\)

Instantaneous current electricity \(\mathbf{i\;=\;lim_{\Delta t\;\rightarrow \;0}\;\frac{\Delta \;Q}{\Delta \;t}\;=\;\frac{dQ}{dt}}\)

If the flow of charge is uniform then \(\mathbf{i\;=\;\frac{Q}{t}}\)

Note: In a conductor, the electric current is caused by the movement of free electrons. The total number of electrons (- ve charge) and protons (+ ve charge) in a conductor is always same. Hence the net charge in a conductor carrying current is zero. Also, the value of current does not change with the change in the cross-sectional area of the conductor.

Electric Current DC
Electric Current AC

Electric Current (i) due to the rotatory motion of the charge (q):

If a point charge (q) is moving with speed v in a circular path of radius r then,

\(\mathbf{i\;=\;q\;v\;=\;\frac{q}{T}\;=\;\frac{q\; u }{2\;\;\pi \;r}\;=\;\frac{q\;\omega }{2\;\pi }}\)

Where, \(\mathbf{\omega }\) = angular velocity, T = time period and \(\mathbf{ u }\) = frequency.

Current electricity when the voltage (V) is applied across the resistance (R):

\(\mathbf{i\;=\;\frac{V}{R}}\) and \(\mathbf{i\;=\;\frac{P}{V}}\) [Where P = Power]

Note: In solids, the current carriers are the free electrons. In liquids, the current carriers are negative and positive ions. In gases, the current carriers are free electrons and positive ions. In semiconductors, the current carriers are holes and free electrons.

Current Density:

Case 1: If the area is normal to the direction of the flow of charge. The Current density (J) at any point is given by:

\(\mathbf{\overrightarrow{J}\;=\;\frac{di}{dA}\;\overrightarrow{n}}\)

Case 2: If the area is normal to the direction of the flow of charge. The Current density (J) at any point is given by:

\(\mathbf{\overrightarrow{J}\;=\;\frac{di}{dA}\;\overrightarrow{n}}\)

Or, \(\mathbf{di\;=\;J\;dA\;cos\; heta \;=\;\overrightarrow{J}\;.\;\overrightarrow{dA}}\)

Therefore, the Relationship between current electricity and current density if the current density is uniform for a normal cross-section:

\(\mathbf{i\;=\int \;\overrightarrow{J}\;.\;\overrightarrow{dA}}\)

Since, \(\mathbf{\overrightarrow{J}}\) = Constant

Therefore, \(\mathbf{i\;=\int \;\overrightarrow{J}\;.\;\overrightarrow{dA}\;=\;\overrightarrow{J}\;.\;\int \;\overrightarrow{ds}\;=\;\overrightarrow{J}\;.\;\overrightarrow{A}\;=\;J A\;cos\;0\;=\;J A }\)

The dimension of Current Density is\(\mathbf{\left [ L^{-2} A \right ]}\)
and SI unit is\(\mathbf{Ampere\;m^{-2}}\)

Different Circuit conditions:

Case 1: When the cell is discharging i.e. closed circuit

A. The Current given by the cell \(\mathbf{i\;=\;\frac{E}{R\;+\;r} }\)

B. The Potential difference (PD) across the resistance is given by V = iR

C. Potential drop inside the cell = ir

D. The Equation of the cell E = V + ir (E > V)

E. The internal resistance (r) of the cell is given by r = \(\mathbf{r\;=\;\left [ \frac{E}{V}\;-\;1 \right ]\;.\;R}\)

F. The Power (P) dissipated in the external resistance (R) is given by:

\(\mathbf{P\;=\;V\;i\;=\;i^{2}\;R\;=\;\frac{V^{2}}{R}\;=\;\left [ \frac{E}{R\;+\;r} \right ]^{2}\;.\;R}\)

G. Power = maximum when R = r i.e. \(\mathbf{P_{max}\;=\;\frac{E^{2}}{4\;r}}\)

Electric Current IIT JEE

Case 2: Open Circuit Condition

A. Current through the circuit (i) = 0

B. Potential difference between A and B, (VB) = E

C. Potential difference between C and D, (VCD) = 0

Case 3: Short Circuit Condition

A. Potential Difference (V) = 0

B. Short circuit current = \(\mathbf{i_{sc}\;=\;\frac{E}{r}}\)

Grouping of Cell:

1. If n identical cells are in series connection

Electric Current IIT JEE

A. The Equiv. EMF Eeq = n E

B. The Equiv. internal resistance req = n r

C. Electric Current from each cell = Main current \(\mathbf{i\;=\;\frac{n\;E}{R\;+\;nr}}\)

D. The Potential difference (PD) across external resistance V = i R

E. The Potential difference (PD) across each cell \(\mathbf{V^{‘}\;=\;\frac{V}{n}}\)

F. Power (P) dissipated in the circuit \(\mathbf{P\;=\;\left [ \frac{n\;E}{R\;+\;nr} \right ]^{2}\;.\;R}\)

G. For maximum power: R = nr

H. Maximum Power \(\mathbf{P_{max}\;=\;n\;\left [ \frac{E^{2}}{4\;r} \right ]}\)

2. If n identical cells are in parallel connection:

Electric Current IIT JEE

A. The Eq. EMF Eeq = E

B. The Eq. internal resistance req = \(\mathbf{\frac{r}{n}}\)

C. Main current \(\mathbf{i\;=\;\frac{E}{R\;+\;\frac{r}{n}}}\)

D. The Potential difference (PD) across external resistance = Potential difference across each cell = V = i R

E. Current electricity from each cell = \(\mathbf{i^{‘}\;=\;\frac{i}{n}}\)

F. Power (P) dissipated in the circuit \(\mathbf{P\;=\;\left [ \frac{E}{R\;+\;\frac{r}{n}} \right ]^{2}\;.\;R}\)

G. For maximum power: \(\mathbf{R\;=\;\frac{r}{n}}\)

H. Maximum Power \(\mathbf{P_{max}\;=\;n\;\left [ \frac{E^{2}}{4\;r} \right ]}\)

Mixed Grouping: If n number of identical cells are connected in series and m rows are in parallel connection as shown:

Electric Current IIT JEE

A. The Equiv. EMF Eeq = n E

B. The Equiv. internal resistance (req) of the combination req = \(\mathbf{\frac{nr}{m}}\)

C. Main current through the load \(\mathbf{i\;=\;\frac{nE}{R\;+\;\frac{nr}{m}}\;=\;\frac{mnE}{mR\;+\;nr}}\)

D. Potential difference (PD) across load V = iR

E. Potential Difference (PD) across each cell \(\mathbf{V’\;=\;\frac{V}{n}}\)

F. Electric Current from each cell \(\mathbf{i’\;=\;\frac{i}{n}}\)

G. For maximum power: \(\mathbf{R\;=\;\frac{nr}{m}}\)

H. Maximum Power \(\mathbf{P_{max}\;=\;mn\left [ \frac{E^{2}}{4r} \right ]}\)

I. Total number of cell = m × n


Practise This Question

We can draw a triangle ABC in which BC is 12cm, angle B is 75 and AB + AC = 6.5 cm.