Numerical Problems Based on Ohms law, Resistivity, Conductivity, Current Density and Colour Code of Carbon Resistors for Class 12 Physics

1. Decode the Resistance:
Using the colour code sequence (BBROYGBVGW):

• First ring (Blue) = $6$
• Second ring (Black) = $0$
• Third ring (Yellow) = Multiplier of $10^4$

The resistance is $R = 60 \times 10^4\ \Omega$.

2. Calculate the Current:
Voltage, $V = 30\text{ V}$. Using Ohm’s Law ($I = \frac{V}{R}$):

Voltage, $V = 10\text{ V}$; Resistance, $R = 1\text{ k}\Omega = 1000\ \Omega$.
Time, $t = 5\text{ minutes} = 5 \times 60 = 300\text{ s}$.

1. Find the current ($I$):

2. Find the total charge ($q$):

3. Calculate the number of electrons ($n$):
Using $q = ne$ (where $e = 1.6 \times 10^{-19}\text{ C}$):

Area, $A = 0.01\text{ mm}^2 = 0.01 \times 10^{-6}\text{ m}^2 = 10^{-8}\text{ m}^2$.
Resistance, $R = 1\text{ k}\Omega = 1000\ \Omega$.
Resistivity, $\rho = 1.7 \times 10^{-8}\ \Omega\text{m}$.

Using the formula for resistance $R = \rho \frac{L}{A}$, rearrange to solve for length ($L$):

Specific resistance (resistivity), $\rho = 64 \times 10^{-8}\ \Omega\text{m}$.
Length, $L = 1.98\text{ m}$. Resistance, $R = 7\ \Omega$.

First, find the cross-sectional area ($A$) using $R = \rho \frac{L}{A}$:

The area of a circle is $A = \pi r^2$. Using $\pi \approx \frac{22}{7}$:

Length, $L = 2\text{ m}$. Resistivity, $\rho = 15 \times 10^{-6}\ \Omega\text{m}$.
Radius, $r = 0.321\text{ mm} = 0.321 \times 10^{-3}\text{ m}$.

1. Calculate the area ($A$):

2. Calculate the resistance ($R$):

3. Calculate the current ($I$):
Voltage, $V = 10\text{ V}$.

Area, $A = 1.0\text{ mm}^2 = 1.0 \times 10^{-6}\text{ m}^2$.
Number of electrons per second, $\frac{n}{t} = 6 \times 10^{16}\text{ s}^{-1}$.
Charge of an electron, $e = 1.6 \times 10^{-19}\text{ C}$.

(i) Calculate the current ($I$):

(ii) Calculate the current density ($J$):

Area, $A = 2.0\text{ mm}^2 = 2.0 \times 10^{-6}\text{ m}^2$. Current, $I = 1\text{ A}$.
Resistivity, $\rho = 1.7 \times 10^{-8}\ \Omega\text{m}$.

From the microscopic form of Ohm’s law, $E = \rho J$, where $J$ is the current density.

1. Calculate current density ($J$):

2. Calculate Electric Field ($E$):

Let the initial diameter be $d$, area be $A$, and length be $L$. Original resistance $R = \rho \frac{L}{A}$.

When the wire is stretched, its volume ($V = A \cdot L$) remains constant.

The new diameter $d’ = \frac{d}{2}$. Because area is proportional to the square of the diameter ($A \propto d^2$), the new area $A’$ is:

Since volume is constant ($A’L’ = AL$), the new length $L’$ must be:

Now, calculate the new resistance ($R’$):

Let the original resistance be $R = \rho \frac{L}{A}$, where $A = \pi r^2$.

The new radius is $r’ = \frac{r}{2}$. The new area $A’$ is:

The new length is explicitly given as $L’ = \frac{L}{4}$. (Note: This is not a stretching problem; the wire’s dimensions are independently modified.)

Calculate the new resistance ($R’$):

Therefore, there is no change in the resistance.

Original resistance, $R = 5\ \Omega$.

When a wire is uniformly stretched, its volume ($A \cdot L$) remains constant. If the new length is $n$ times the original length ($L’ = nL$), the new area becomes $A’ = \frac{A}{n}$ to maintain the volume.

Here, $n = 4$, so $L’ = 4L$ and $A’ = \frac{A}{4}$.

The new resistance ($R’$) is given by:

This shows $R’ = n^2 R = 4^2 R = 16R$.