Numerical Problems-1 - Chapter 2 Current Electricity 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Numerical Problems
Question 1.
The following graphs represent the current versus voltage and voltage versus current for the six conductors A,B,C,D,E and F. Which conductor has least resistance and which has maximum resistance?

Solution:

According to ohm's law, $V=I R$
Resistance of conductor, $\mathrm{R}=\frac{V}{I}$
Graph-I:
Conductor $\mathrm{A}, \mathrm{I}=4 \mathrm{~A}$ and $\mathrm{V}=2 \mathrm{~V}$
$
\mathrm{R}=\frac{V}{I}=\frac{2}{4}=0.5 \Omega
$
Conductor $\mathrm{B}, \mathrm{I}=3 \mathrm{~A}$ and $\mathrm{V}=4 \mathrm{~V}$
$
\mathrm{R}=\frac{V}{I}=\frac{4}{3}=1.33 \Omega
$
Conductor $\mathrm{C}, \mathrm{I}=2 \mathrm{~A}$ and $\mathrm{V}=5 \mathrm{~V}$
$
\mathrm{R}=\frac{V}{I}=\frac{5}{2}=2.5 \mathrm{Q} \Omega
$
Graph-II:
Conductor $\mathrm{D}, \mathrm{I}=2 \mathrm{~A}$ and $\mathrm{V}=4 \mathrm{~V}$
$
\mathrm{R}=\frac{V}{I}=\frac{4}{2}=2 \Omega \mathrm{d}
$
Conductor $\mathrm{E}, \mathrm{I}=4 \mathrm{~A}$ and $\mathrm{V}=3 \mathrm{~V}$
$
\mathrm{R}=\frac{V}{I}=\frac{3}{4}=1.75 \Omega
$
Conductor $\mathrm{F}, \mathrm{I}=5 \mathrm{~A}$ and $\mathrm{V}=2 \mathrm{~V}$
$
\mathrm{R}=\frac{V}{I}=\frac{2}{5}=0.4 \Omega
$
Conductor $F$ has least resistance, $\mathrm{R}_{\mathrm{F}}=0.4 \Omega$,
Conductor $\mathrm{C}$ has maximum resistance, $\mathrm{R}_{\mathrm{C}}=2.5 \mathrm{G}$.
Question 2.
Lightning is very good example of natural current. In typical lightning, there is $10^9 \mathrm{~J}$ energy transfer across the potential difference of $5 \times 10^7 \mathrm{~V}$ during a time interval of $0.2 \mathrm{~s}$.

Using this information, estimate the following quantities (a) total amount of charge transferred between cloud and ground (b) the current in the lightning bolt (c) the power delivered in $0.2 \mathrm{~s}$.
Solution:
During the lightning energy, $\mathrm{E}=10^9 \mathrm{~J}$
Potential energy, $\mathrm{V}=5 \times 10^7 \mathrm{~V}$
Time interval, $\mathrm{t}=0.2 \mathrm{~s}$
(a) Amount of charge transferred between cloud and ground, $\mathrm{q}=\mathrm{It}$
(b) Current in the lighting bolt, $\mathrm{E}=$ VIt
$
\begin{aligned}
& \mathrm{I}=\frac{E}{V t}=\frac{10^9}{5 \times 10^7 \times 0.2}=1 \times 10^9 \times 1^{-7} \\
& \mathrm{I}=1 \times 10^2 \\
& \mathrm{I}=100 \mathrm{~A} \\
& \therefore \mathrm{q}=\mathrm{It}=100 \times 0.2 \\
& \mathrm{q}=20 \mathrm{C} .
\end{aligned}
$
(c) Power delivered, $\mathrm{E}=$ VIt
$
\begin{aligned}
& \mathrm{P}=\mathrm{VI}=5 \times 10^7 \times 100=500 \times 10^7 \\
& \mathrm{I}=5 \times 10^9 \mathrm{~W} \\
& \mathrm{P}=5 \mathrm{GW}
\end{aligned}
$

Question 3.
A copper wire of $10^6 \mathrm{~m}^2$ area of cross section, carries a current of $2 \mathrm{~A}$. If the number of electrons per cubic meter is $8 \times 10^{28}$, calculate the current density and average drift velocity.
Solution:
Cross-sections area of copper wire, $\mathrm{A}=10^6 \mathrm{~m}^2$ $\mathrm{I}=2 \mathrm{~A}$
Number of electron, $\mathrm{n}=8 \times 10^{28}$
Current density, $\mathrm{J}=\frac{1}{A}=\frac{2}{10^{-6}}$
$\mathrm{J}=2 \times 10^6 \mathrm{Am}^{-2}$
Average drift velocity, $\mathrm{V}_{\mathrm{d}}=\frac{1}{n e A}$
$\mathrm{e}$ is the charge of electron $=1.6 \times 10^{-9} \mathrm{C}$
$\mathrm{V}_{\mathrm{d}}=\frac{2}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}}=\frac{1}{64 \times 103}$
$\mathrm{V}_{\mathrm{d}}=0.15625 \times 10^{-3}$
$\mathrm{V}_{\mathrm{d}}=15.6 \times 10^{-5} \mathrm{~ms}^{-1}$
Question 4.
The resistance of a nichrome wire at $0{ }^{\circ} \mathrm{C}$ is $10 \Omega$. If its temperature coefficient of resistance is $0.004 /{ }^{\circ} \mathrm{C}$, find its resistance at boiling point of water. Comment on the result.
Solution:
Resistance of a nichrome wire at $0^{\circ} \mathrm{C}, \mathrm{R}_0=10 \Omega$
Temperature co-efficient of resistance, $\alpha=0.004 /{ }^{\circ} \mathrm{C}$
Resistance at boiling point of water, $\mathrm{RT}=$ ?
Temperature of boiling point of water, $\mathrm{T}=100^{\circ} \mathrm{C}$
$\mathrm{R}_{\mathrm{T}}=\mathrm{R}_0(1+\alpha \mathrm{T})=10[1+(0.004 \times 100)]$
$\mathrm{R}_{\mathrm{T}}=10(1+0.4)=10 \times 1.4$
$\mathrm{R}_{\mathrm{T}}=14 \Omega$
As the temperature increases the resistance of the wire also increases.
Question 5.
The rod given in the figure is made up of two different materials.

Both have square cross sections of $3 \mathrm{~mm}$ side. The resistivity of the first material is $4 \times 10^{-3} \Omega . \mathrm{m}$ and it is $25 \mathrm{~cm}$ long while second material has resistivity of $5 \times 10^{-3} \Omega . \mathrm{m}$ and is of $70 \mathrm{~cm}$ long. What is the resistivity of rod between its ends?
Solution:
Square cross section of side, $\mathrm{a}=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}$
Cross section of side, $A=a^2=9 \times 10^6 \mathrm{~m}$
First material:
Resistivity of the material, $\rho_1=4 \times 10^{-3} \Omega \mathrm{m}$
length, $1_1=25 \mathrm{~cm}=25 \times 10^{-2} \mathrm{~m}$
Resistance of the lord, $\mathrm{R}_1=\frac{\rho_l l_l}{A}=\frac{4 \times 10^{-3} \times 25 \times 10^{-2}}{9 \times 10^{-6}}=\frac{100 \times 10^{-6} \times 10^6}{9}$
$
\mathrm{R}_1=11.11 \times 10^1 \Omega
$
Second material:
Resistivity of the material, $\rho_2=5 \times 10^{-3} \Omega \mathrm{m}$
length, $1_2=70 \mathrm{~cm}=70 \times 10^{-2} \mathrm{~m}$
Resistance of the rod, $\mathrm{R}_2=\frac{\rho_2 l_2}{A}=\frac{5 \times 10^{-3} \times 70 \times 10^{-2}}{9 \times 10^{-6}}=\frac{350 \times 10^{-6} \times 10^6}{9}$
$
\mathrm{R}_2=38.88 \times 10^1
$
$
\mathrm{R}_2=389 \Omega
$
Total resistance between the ends $\mathrm{f}$ the rods
$
\begin{aligned}
& \mathrm{R}=\mathrm{R}_1+\mathrm{R}_2=111+389 \\
& =500 \Omega
\end{aligned}
$
Question 6.
Three identical lamps each having a resistance $\mathrm{R}$ are connected to the battery of emf as shown in the figure.

Suddenly the switch S is closed, (a) Calculate the current in the circuit when S is open and closed (b) What happens to the intensities of the bulbs A,B and C. (c) Calculate the voltage across the three bulbs when $\mathrm{S}$ is open and closed (d) Calculate the power delivered to the circuit when $\mathrm{S}$ is opened and closed (e) Does the power delivered to the circuit decreases, increases or remain same?
Solution:
Resistance of the identical $\mathrm{lamp}=\mathrm{R}$

Emf of the battery $=\xi$
According to Ohm's Law, $\zeta=$ IR
(a) Current:
When Switch is open-The current in the circuit. Total resistance of the bulb,
$
\begin{aligned}
& \mathrm{R}_{\mathrm{s}}=\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3 \\
& \mathrm{R}_1=\mathrm{R}_2=\mathrm{R}_3=\mathrm{R} \\
& \mathrm{R}_{\mathrm{s}}=\mathrm{R}+\mathrm{R}+\mathrm{R}=3 \mathrm{R} \\
& \therefore \text { Current, } \mathrm{I}=\frac{\xi}{R_s} \\
& \Rightarrow \mathrm{I}_0=\frac{\xi}{3 R}
\end{aligned}
$
Switch is closed- The current in the circuit. Total resistance of the bulb, $\mathrm{R}_{\mathrm{S}}=\mathrm{R}+\mathrm{R}=2 \mathrm{R}$
Current $\mathrm{I}=\frac{\xi}{R_9}$
$
\mathrm{I}_{\mathrm{c}}=\frac{\xi}{2 R}
$

(b) Intensity:
When switch is open - All the bulbs glow with equal intensity.
When switch is closed - The intensities of the bulbs A and B equally increase. Bulb C will not glow since no current pass through it.
(c) Voltage across three bulbs:
When switch is open - Voltage across bulb A, $\mathrm{V}_{\mathrm{A}}=\mathrm{I}_0 \mathrm{R}=\frac{\xi}{3 R} \times \mathrm{R}=\frac{\xi}{3}$ similarly:
Voltage across bulb $\mathrm{B}, \mathrm{V}_{\mathrm{B}}=\frac{\xi}{3}$
Voltage across bulb C, $\mathrm{V}_{\mathrm{C}}=\frac{\xi}{3}$
When switch is closed-Voltage across bulb A, $\mathrm{V}_{\mathrm{A}}=\mathrm{I}_{\mathrm{C}} \mathrm{R}=\frac{\xi}{2 R} \times \frac{\xi}{2}$ similarly:
Voltage across bulb $\mathrm{B}, \mathrm{V}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}} \mathrm{R} \frac{\xi}{2}$
Voltage across bulb C, $\mathrm{V}_{\mathrm{C}}=0$
(d) Power delivered to the circuit,
When switch is opened - Power $\mathrm{P},=\mathrm{VI}$

$
\mathrm{P}_{\mathrm{A}}=\mathrm{V}_{\mathrm{A}} \mathrm{I}_0=\frac{\xi}{3} \times \frac{\xi}{3 \mathrm{R}}=\frac{\xi^2}{9 \mathrm{R}}
$
Similarly: $\quad P_B=\frac{\xi^2}{9 R}$ and $P_c=\frac{\xi^2}{9 R}$
When switch is closed - Power $\mathrm{P},=\mathrm{VI}$
$
\mathrm{P}_{\mathrm{A}}=\mathrm{V}_{\mathrm{A}} \mathrm{I}_c=\frac{\xi}{2} \times \frac{\xi}{2 \mathrm{R}}=\frac{\xi^2}{4 \mathrm{R}}
$
Similarly: $\mathrm{P}_{\mathrm{B}}=\frac{\xi^2}{4 \mathrm{R}}$ and $\mathrm{P}_c=0$
(e) Total power delivered to circuit increases.

Question 7.
The current through an element is shown in the figure. Determine the total charge that pass through the element at
(a) $t=0 s$
(b) $t=2 s$
(c) $t=5 s$

Solution :
Rate of flow of charge is called current, $\mathrm{I}=\frac{d q}{d t}$.
Total charge pass through element, $\mathrm{dq}=\mathrm{Idt}$
(a) $\mathrm{At} \mathrm{t}=0 \mathrm{~s}, \mathrm{I}=10 \mathrm{~A}$
$\mathrm{dq}=\mathrm{Idt}=10 \times 0=0 \mathrm{C}$.
(b) At t $=2 \mathrm{~s}, \mathrm{I}=5 \mathrm{~A}$
$\mathrm{dq}=\mathrm{Idt}=5 \times 2=10 \mathrm{C}$.
(c) At $t=5 \mathrm{~s}, \mathrm{I}=0$
$\mathrm{dq}=\mathrm{Idt}=0 \times 5=0 \mathrm{C}$.
Question 8.
An electronics hobbyist is building a radio which requires $150 \Omega$ in her circuit, but she has only $220 \Omega$, $79 \Omega$ and $92 \Omega$ resistors available. How can she connect the available resistors to get desired value of resistance?
Solution:
Required effective resistance $=150 \Omega$
Given resistors of resistance, $\mathrm{R}_1=220 \Omega, \mathrm{R}_2=79 \Omega, \mathrm{R}_3=92 \Omega$
Parallel combination $\mathrm{R}_1$ and $\mathrm{R}_2$
$
\begin{aligned}
& \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{220}+\frac{1}{79}=\frac{79+220}{220 \times 79} \\
& \mathrm{R}_{\mathrm{p}}=58 \Omega
\end{aligned}
$

Parallel combination $\mathrm{R}_{\mathrm{p}}$ and $\mathrm{R}_3$
$
\begin{aligned}
& \mathrm{R}_{\mathrm{s}}=\mathrm{R}_{\mathrm{p}}+\mathrm{R}_3=58+92 \\
& \mathrm{R}_{\mathrm{s}}=150 \Omega
\end{aligned}
$
Parallel combination of $220 \Omega$ and $79 \Omega$ in series with $92 \Omega$.
Question 9.
A cell supplies a current of 0.9 A through a $2 \Omega$ resistor and a current of $0.3 \mathrm{~A}$ through a $7 \Omega$ resistor. Calculate the internal resistance of the cell.
Solution:
Current from the cell, $\mathrm{I}_1=0.9 \mathrm{~A}$
Resistor, $\mathrm{R}_1=2 \Omega$
Current from the cell, $\mathrm{I}_2=0.3 \mathrm{~A}$
Resistor, $\mathrm{R}_2 7 \Omega$
Internal resistance of the cell, $\mathrm{r}=$ ?
Current in the circuit $\mathrm{I}_1=\frac{\xi}{r+R_1}$
$\xi=\mathrm{I}_1\left(\mathrm{r}+\mathrm{R}_1\right) \ldots . .(1)$
Current in the circuit, $\mathrm{I}_2=\frac{\xi}{r+R_2}$
Equating equation (1) and (2),
$\mathrm{I}_1 \mathrm{r}+\mathrm{I}_1 \mathrm{R}_1=\mathrm{I}_2 \mathrm{R}_2+\mathrm{I}_2 \mathrm{r}$
$\left(\mathrm{I}_1-\mathrm{I}_2\right) \mathrm{r}=\mathrm{I}_2 \mathrm{R}_2-\mathrm{I}_1 \mathrm{R}_1$
$r=\frac{\mathrm{I}_2 \mathrm{R}_2-\mathrm{I}_1 \mathrm{R}_1}{\mathrm{I}_1-\mathrm{I}_2}=\frac{(0.3 \times 7)-(0.9 \times 2)}{0.9-0.3}=\frac{2.1-1.8}{0.6}=\frac{0.3}{0.6}$
$\mathrm{r}=0.5 \Omega$.
Question 10.
Calculate the currents in the following circuit.

Solution:
Applying Kirchoff's $1^{\text {st }}$ Law at junction B

$
\begin{aligned}
& \mathrm{I}_1-\mathrm{I}_1-\mathrm{I}_3=0 \\
& \mathrm{I}_3=\mathrm{I}_1-\mathrm{I}_2 \ldots .
\end{aligned}
$
Applying Kirchoff's II ${ }^{\text {nd }}$ Law at junction in ABEFA
$
\begin{aligned}
& 100 I_3+100 I_1=15 \\
& 100\left(I_3+I_1\right)=15 \\
& 100 I_1-100 I_2+100 I_1=15 \\
& 200 I_1-100 I_2=15 \ldots \ldots \text { (2) }
\end{aligned}
$
Applying Kirchoff's IInd Law at junction in BCDED
$
\begin{aligned}
& -100 \mathrm{I}_2+100 \mathrm{I}_3=9 \\
& -100 \mathrm{I}_2+100\left(\mathrm{I}_1-\mathrm{I}_2\right)=9 \\
& 100 \mathrm{I}_1-200 \mathrm{I}_2=9
\end{aligned}
$
Solving equating (2) and (3)

Substitute $I_1$ values in equ (2)
$
\begin{aligned}
& 200(0.07)-100 \mathrm{I}_2=15 \\
& 14-100 \mathrm{I}_2=15 \\
& -100 \mathrm{I}_2=15-14 \\
& \mathrm{I}_2=\frac{-1}{100} \\
& \mathrm{I}_2=-0.01 \mathrm{~A}
\end{aligned}
$
Substitute $I_1$ and $I_2$ value in equ (1), we get
$
\begin{aligned}
& \mathrm{I}_3=\mathrm{I}_1-\mathrm{I}_2=0.07-(-0.01) \\
& \mathrm{I}_3=0.08 \mathrm{~A} .
\end{aligned}
$
Question 11
A potentiometer wire has a length of $4 \mathrm{~m}$ and resistance of $20 \Omega$. It is connected in series with resistance of $2980 \Omega$ and a cell of emf $4 \mathrm{~V}$. Calculate the potential along the wire.
Solution:
Length of the potential wire, $1=4 \mathrm{~m}$
Resistance of the wire, $r=20 \Omega$
Resistance connected series with potentiometer wire, $R=2980 \Omega$
Emf of the cell, $\xi=4 \mathrm{~V}$
Effective resistance, $\mathrm{R}_{\mathrm{S}}=\mathrm{r}+\mathrm{R}=20+2980=3000 \Omega$
Current flowing through the wire, $\mathrm{I}=\frac{\xi}{r+R_8}=\frac{4}{3000}$
$\mathrm{I}=1.33 \times 10^{-3} \mathrm{~A}$
Potential drop across the wire, $\mathrm{V}=\mathrm{Ir}$
$
\begin{aligned}
& =1.33 \times 10^{-3} 3 \times 20 \\
& \mathrm{~V}=26.6 \times 10^{-3} \mathrm{~V}
\end{aligned}
$

Potential gradient $=\frac{\text { Potential drop across the wire }}{\text { length of the wire }}=\frac{\mathrm{V}}{l}=\frac{26.6 \times 10^{-3}}{4}$ $=6.65 \times 10^{-3}$
Potential garadient $=0.66 \times 10^{-2} \mathrm{Vm}^{-1}$
Question 12.
Determine the current flowing through the galvanometer (G) as shown in the figure.

Solution:
Current flowing through the circuit, $I=2 \mathrm{~A}$
Applying Kirchoff's I $\mathrm{st}^{\text {st }}$ law at junction $\mathrm{P}, \mathrm{I}=\mathrm{I}_1+\mathrm{I}_2$ $\mathrm{I}_2=\mathrm{I}-\mathrm{I}_1 \ldots$ (1)
Applying Kirchoff's Ind law at junction PQSP $5 \mathrm{I}_1+10 \mathrm{I}_{\mathrm{g}}-15 \mathrm{I}_2=0$ $5 \mathrm{I}_1+10 \mathrm{I}_{\mathrm{g}}-15\left(\mathrm{I}-\mathrm{I}_1\right)=0$ $20 \mathrm{I}_1+10 \mathrm{I}_{\mathrm{g}}=15 \mathrm{I}$ $20 \mathrm{I}_1+10 \mathrm{I}_{\mathrm{g}}=15 \times 2$ $\div$ by $1021 \mathrm{I}_1+\mathrm{I}_{\mathrm{g}}=3 \ldots$ (2)
Applying Kirchoff's $\mathrm{II}^{\text {nd }}$ law at junction QRSQ $10\left(I_1-I_g\right)-20\left(I-I_1-I_g\right)-10 I_g=0$ $10 \mathrm{I}_1-10_{\mathrm{g}}-20\left(\mathrm{I}-\mathrm{I}_1-\mathrm{I}_{\mathrm{g}}\right)-10 \mathrm{I}_{\mathrm{g}}=0$ $10 \mathrm{I}_1-10 \mathrm{~g}-20 \mathrm{I}+20 \mathrm{I}_1-20 \mathrm{I}_{\mathrm{g}}-10 \mathrm{I}_{\mathrm{g}}=0$ $30 \mathrm{I}_1-40 \mathrm{I}_{\mathrm{g}}=20 \mathrm{I}$
$\div$ by $10 \Rightarrow 3 \mathrm{I}_1-4 \mathrm{I}_{\mathrm{g}}=20 \mathrm{I}$
$3 \mathrm{I}_1-4 \mathrm{I}_{\mathrm{g}}=2 \mathrm{I}$
$3 I_1-4 I_g=2 \times 2=4$
equ $(2) \times 3 \quad 6 \mathrm{I}_1+3 \mathrm{I}_{\mathrm{g}}=9$
equ $(3) \times 2 \quad 6 I_1-8 I_{\mathrm{g}}=8$