Ohm's Law

Current Electricity · CBSE

Theory

Statement and the meaning of $R$

Ohm's law: at constant temperature the current $I$ through a conductor is proportional to the potential difference $V$ across its ends,

$V \propto I \qquad \Rightarrow \qquad V = IR$

The constant $R$ is the resistance of the conductor. SI unit ohm, $1\ \Omega = 1\ \text{V A}^{-1}$ .
Careful with the wording CBSE marks: $V = IR$ only defines resistance, and can be written for any conducting device whether it obeys Ohm's law or not. Ohm's law is the further assertion that the $I$ – $V$ plot is a straight line through the origin, i.e. that $R$ is independent of $V$ .

Deriving $R = \rho l / A$ (NCERT's slab argument)

Take a conductor as a slab of length $l$ and cross-sectional area $A$ , obeying $V = IR$ .
Put two identical slabs side by side end to end. The same current $I$ flows through the combination as through either slab, and the potential difference across each is $V$ , so the total is $2V$ across a length $2l$ . Hence

$R_C = \dfrac{2V}{I} = 2R \qquad \Rightarrow \qquad R \propto l$

Now cut the same slab lengthwise into two, each of length $l$ but area $A/2$ . For the same $V$ across the slab, each half carries $I/2$ , so each half has

$R_1 = \dfrac{V}{I/2} = 2\dfrac{V}{I} = 2R \qquad \Rightarrow \qquad R \propto \dfrac{1}{A}$

Combining, $R \propto l/A$ , and for a given conductor

$R = \dfrac{\rho l}{A}$

The constant $\rho$ depends on the material, not on the dimensions, and is called the resistivity. Its SI unit is $\Omega$ m.

The point (microscopic) form

Substituting into Ohm's law, $V = I\rho l / A$ . Write $j = I/A$ for the current density and $E = V/l$ for the uniform field in the conductor of length $l$ . Then $El = j\rho l$ , so

$E = \rho j \qquad \text{and, in vector form,} \qquad \mathbf{E} = \rho\,\mathbf{j}, \qquad \mathbf{j} = \sigma \mathbf{E}$

Here $\sigma = 1/\rho$ is the conductivity. This is the equivalent statement of Ohm's law: a material obeys Ohm's law when its resistivity does not depend on the magnitude or direction of the applied field.

Conductivity — the 1-mark definition

Electrical conductivity $\sigma$ of a material is the reciprocal of its resistivity, $\sigma = 1/\rho = j/E$ . It measures the ease with which the material lets current flow.
SI unit: $\Omega^{-1}\text{m}^{-1}$ , also written mho m $^{-1}$ or siemens per metre, S m $^{-1}$ .
Ratio $j/E$ = conductivity; ratio $E/j$ = resistivity. CBSE has set both directions as MCQs.

Reshaping a conductor

Stretching keeps the volume $Al$ constant. Stretch to $n$ times the length and the area falls to $A/n$ , so

$R' = \dfrac{\rho(nl)}{A/n} = n^{2}R$

Cutting a rod of length $l$ and area $A$ lengthwise into two identical pieces and joining them end to end gives length $2l$ and area $A/2$ :

$R' = \dfrac{\rho(2l)}{A/2} = \dfrac{4\rho l}{A} = 4R$

So to keep the same current $I$ flowing, the potential difference must be raised to $V' = IR' = 4V$ — CBSE's Foreign 2016 answer exactly.
Two wires of the same material in parallel with lengths in the ratio $3:2$ and radii in the ratio $2:3$ : since $R = \rho l / \pi r^2$ , $R_1/R_2 = (3/2)(3/2)^2 = 27/8$ , and in parallel $I \propto 1/R$ , so $I_1 : I_2 = 8 : 27$ .

Non-uniform cross-section

In the steady state charge cannot accumulate anywhere, so the same current crosses every section: the current $I$ is the quantity that stays constant along a wire of varying thickness.
Since $I = jA = neAv_d = EA/\rho$ , each of $j$ , $v_d$ and $E$ varies as $1/A$ : all three are larger where the wire is thinner and smaller where it is thicker.
Two wires of the same material and the same length, of areas $A$ and $2A$ , connected in parallel across an ideal battery of emf $E$ : both have the same $V$ , hence the same field $E_{\text{field}} = V/l$ , hence the same $j = \sigma E_{\text{field}}$ . So $j_x : j_y = 1 : 1$ , independent of area.

Standard results

$V = IR$ ; $R = \rho l/A$ ; $\rho = RA/l$ ; $\sigma = 1/\rho$ .
$j = I/A$ ; $E = V/l$ ; $E = \rho j$ ; $j = \sigma E$ .
Stretched to $n$ times the length: $R' = n^2R$ . Cut lengthwise and rejoined end to end: $R' = 4R$ .
Along a conductor of non-uniform area, only $I$ is constant; $j$ , $v_d$ , $E$ all vary as $1/A$ .

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
Two wires $X$ and $Y$ of the same material and of equal lengths having areas of cross-section $A$ and $2A$ respectively, are connected in parallel across an ideal battery of emf $E$ . What is the ratio of current density $(j_x/j_y)$ in them?
1 markCBSE Class 12 Physics — Very Short Answer · 2021C · Section A, 1 markAnswer 1 ↓
2
A copper wire of non-uniform area of cross-section is connected to a d.c. battery. The physical quantity which remains constant along the wire is ______.
1 markCBSE Class 12 Physics — Very Short Answer · 2020, R · Section A, 1 markAnswer 2 ↓
3
Define the term 'electrical conductivity' of a metallic wire. Write its S.I. unit.
1 markCBSE Class 12 Physics — Very Short Answer · AI 2017C, Delhi 2014, R · Section A, 1 markAnswer 3 ↓
4
A metal rod of square cross-sectional area $A$ having length $l$ has current $I$ flowing through it when a potential difference of $V$ volt is applied across its ends (figure I). Now the rod is cut parallel to its length into two identical pieces and joined as shown in figure II. What potential difference must be maintained across the length $2l$ so that the current in the rod is still $I$ ?
2 marksCBSE Class 12 Physics — SA · Foreign 2016, Ap · Section B, 2 marksAnswer 4 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

5
A potential difference of 200 V is maintained across a conductor of resistance 100 $\Omega$ . The number of electrons passing through it in 1 s is
- (a)
  $1.25 \times 10^{19}$
- (b)
  $2.5 \times 10^{18}$
- (c)
  $1.25 \times 10^{18}$
- (d)
  $2.5 \times 10^{16}$
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 5 ↓
6
The ratio of current density and electric field is called OR In a current carrying conductor, the ratio of the electric field and the current density at a point is called
- (a)
  resistivity
- (b)
  conductivity
- (c)
  drift velocity
- (d)
  mobility
1 markCBSE Class 12 Physics — Multiple Choice · 2020 · Section A, 1 markAnswer 6 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
Since $X$ and $Y$ are connected in parallel across the same ideal (zero internal resistance) battery, the potential difference across each wire is the same, $= E$ . As both wires have the same length $l$ , the electric field inside each, $E_{field} = V/l$ , is identical. Current density $J = \sigma E_{field}$ depends only on the material's conductivity $\sigma$ and the field $E_{field}$ — both are the same for $X$ and $Y$ — and is independent of the cross-sectional area. $\therefore j_x : j_y = 1 : 1$
2.↑ back to the question
In a steady state, charge cannot accumulate at any point of the conductor, so the same amount of charge must cross every cross-section in a given time. Hence the electric current $I$ remains constant along the wire (even though the area, and therefore current density, drift speed and electric field, all vary along its length).
3.↑ back to the question
Electrical conductivity ( $\sigma$ ) of a material is the reciprocal of its resistivity: $\sigma = 1/\rho = J/E$ . It measures the ease with which the material allows the flow of electric current. SI unit: $\Omega^{-1}\text{m}^{-1}$ (also written mho m $^{-1}$ or siemens/metre, S m $^{-1}$ ).
4.↑ back to the question
Original rod: $V = I\,\dfrac{\rho l}{A}$ ...(i) When the rod is cut parallel to its length into two identical pieces and rejoined end to end (as in figure II), the new conductor has length $2l$ and cross-sectional area $A/2$ . New resistance: $R' = \dfrac{\rho (2l)}{A/2} = \dfrac{4\rho l}{A} = 4R$ For the same current $I$ : $V' = I R' = 4\left(I\dfrac{\rho l}{A}\right) = 4V$ [Using (i)] So the new potential difference required is four times the original potential difference $V$ .
5.Answer: (A)↑ back to the question
Answer: (a) $1.25 \times 10^{19}$ $V = 200$ V, $R = 100\, \Omega$ , $t = 1$ s. $I = \dfrac{V}{R} = \dfrac{200}{100} = 2$ A Charge flowing in 1 s: $q = It = 2 \times 1 = 2$ C Number of electrons, $n = \dfrac{q}{e} = \dfrac{2}{1.6 \times 10^{-19}} = 1.25 \times 10^{19}$
6.Answer: (A)↑ back to the question
Answer: (a) resistivity Since $J = \sigma E$ , the ratio $J/E = \sigma$ , the conductivity of the material. OR Since $J = \sigma E \Rightarrow E/J = 1/\sigma = \rho$ , the resistivity of the material.