Numerical Problems-2 - Chapter 2 Current Electricity 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
Show that one ampere is equivalent to a flow of $6.25 \times 10^{18}$ elementary charges per second.
Solution:
Here $I=1 \mathrm{~A}, \mathrm{t}=1 \mathrm{~s}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$
As $\mathrm{I}=\frac{q}{t}=\frac{n e}{t}$
Number of electrons, $\mathrm{n}=\frac{I t}{e}=\frac{1 \times 1}{1.6 \times 10^{19}}=6.25 \times 10^{18}$
Question 2.
Calculate the resistivity of a material of a wire $10 \mathrm{~m}$ long. $0.4 \mathrm{~mm}$ in diameter and having a resistance of $2.0 \Omega$
Solution:
Here $1=10 \mathrm{~cm}, \mathrm{r}=0.2 \mathrm{~mm}=0.2 \times 10^{-3} \mathrm{~m}, \mathrm{R}=2 \Omega$
$\left[r=\frac{d}{2}\right]$

Resistivity, $l=\frac{\mathrm{RA}}{l}=\frac{\mathrm{R} \times \pi r^2}{l}=\frac{2 \times 3.14 \times\left(0.2 \times 10^{-3}\right)^2}{10}$ $=2.513 \times 10^{-8} \Omega$.

Question 3.
A wire of 10 ohm resistance is stretched to thrice its original length. What will be its (i) new resistivity and (ii) new resistance?
Solution:
(i) Resistivity p remains unchanged because it is the property of the material of the wire.
(ii) In both cases, volume of wire is same, so
$
\begin{aligned}
\mathrm{V} & =\mathrm{A}^{\prime} l^{\prime}=\mathrm{A} l \\
\frac{\mathrm{A}^{\prime}}{\mathrm{A}} & =\frac{l}{l^{\prime}}=\frac{l}{3 l}=\frac{1}{3} \quad\left[\because l^{\prime}=l+2 l=3 l\right] \\
\therefore \quad \frac{\mathrm{R}^{\prime}}{\mathrm{R}} & =\frac{\rho \frac{l^{\prime}}{\mathrm{A}^{\prime}}}{\rho \frac{l}{\mathrm{~A}}}=\frac{l^{\prime}}{l} \times \frac{\mathrm{A}}{\mathrm{A}^{\prime}}=\frac{3}{1} \times \frac{3}{1}=9
\end{aligned}
$
Hence, $\quad \mathrm{R}^{\prime}=9 \mathrm{R}=9 \times 10=90 \Omega$

Question 4.
A copper wire has a resistance of $10 \Omega$. and an area of cross-section $1 \mathrm{~mm}^2$. A potential difference of 10 V exists across the wire. Calculate the drift speed of electrons if the number of electrons per cubic metre in copper is $8 \times 10^{28}$ electrons.
Solution:
Here, $\mathrm{R}=10 \Omega, \mathrm{A}=1 \mathrm{~mm}^2=10^{-6} \mathrm{~m}^2, \mathrm{~V}=10 \mathrm{~V}, \mathrm{n}=8 \times 10^{28}$ electrons $/ \mathrm{m}^3$
Now,
$\mathrm{I}=\mathrm{enAv}_{\mathrm{d}}$
$\therefore \frac{V}{R}=$ enAv $\mathrm{d}_{\mathrm{d}}$ (or) $\mathrm{v}_{\mathrm{d}}=\frac{V}{e n A R}$
$=\frac{10}{1.6 \times 10^{-19} \times 8 \times 10^{23} \times 10^{-6} \times 10}=0.078 \times 10^{-3} \mathrm{~ms}^{-1}$
$=0.078 \mathrm{x} \mathrm{ms}^{-1}$
Question 5.
(i) At what temperature would the resistance of a copper conductor be double its resistance at $0^{\circ} \mathrm{C}$.
(ii) Does this temperature hold for all copper conductors regardless of shape and size? Given a for $\mathrm{Cu}=$ $3.9 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}$.
Solution:

$
\begin{gathered}
\alpha=\frac{\mathrm{R}_2-\mathrm{R}_1}{\mathrm{R}_1\left(t_2-t_1\right)}=\frac{2 \mathrm{R}_0-\mathrm{R}_0}{\mathrm{R}_0(t-0)}=\frac{1}{t} \\
t=\frac{1}{\alpha}=\frac{1}{3.9 \times 10^{-3}}=256^{\circ} \mathrm{C}
\end{gathered}
$
$(i)$
Thus the resistance of copper conductor becomes double at $256^{\circ} \mathrm{C}$.
(ii) Since a does not depend on size and shape of the conductor. So the above result holds for all copper conductors.
Question 6.
Find the value of current $I$ in the circuit shown in figure.
Solution:

In the circuit, the resistance of arm $\mathrm{ACB}(30+30=60 \Omega)$ is the parallel with the resistance of arm $\mathrm{AB}(=$ $30 \Omega)$.
Hence, the effective resistance of the circuit is
$
\mathrm{R}=\frac{30 \times 60}{30+60}=20 \Omega
$
Current, $\mathrm{I}=\frac{V}{R}=\frac{2}{20}=0.1 \mathrm{~A}$.
Common Errors and Its Rectifications:
Common Errors:
1. Sometimes students think that charge and current are same.
2. In doing calculation part students can't give the importance to mention the units.
3. They may confuse the parallel and series network of the resistance.
Rectifications:
1. Charge $q=$ ne Current $I=q / t$
2. Unit is very importance to the every physical quantities.
3. If the resistors are series, their resultant is sum of the all the reciprocal of individual resistance. If the resistors are parallel their resultant is sum of the individual resistance.