Numerical Problems-2 - Chapter 4 Electromagnetic Induction and Alternating Current 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
A coil has 2000 turns and area $70 \mathrm{~cm}^2$. The magnetic field perpendicular to the plane of the coil is 0.3 $\mathrm{Wb} / \mathrm{m}^2$. The coil takes 0.1 s to rotate through $180^{\circ}$. Then what is the value of induced emf?
Solution:
Magnitude of change in flux,
$|\Delta \Phi|=\mid \mathrm{NBA}\left(\cos 180^{\circ}-\cos 0^{\circ}\right.$
$=|\mathrm{NBA}(-1-1)|=|-2 \mathrm{NBA}|=|2 \mathrm{NBA}|$
Where,
$\mathrm{N}=2000$
$\mathrm{B}=0.3 \mathrm{~Wb} / \mathrm{m}^2$
$\mathrm{A}=70 \times 10^{-4} \mathrm{~m}^2$
$\mathrm{t}=0.1 \mathrm{sec}$
Induced emf, $\varepsilon=\frac{|\Delta \phi|}{\Delta t}=\frac{2 N B A}{\Delta t}=\frac{2 \times 2000 \times 0.3 \times 70 \times 10^{-4}}{0.1}$
$
\varepsilon=84 \mathrm{~V}
$

Question 2.
A rectangular loop of sides $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ is lying in a uniform magnetic field of magnitude $0.5 \mathrm{~T}$ with its plane normal to the field. The field is now gradually reduced at the rate of $0.02 \mathrm{~T} / \mathrm{s}$. If the resistance of the loop is $1.6 \Omega$, then find the power dissipated by the loop as heat.
Solution:
Induced emf, $|\varepsilon|=\frac{d \Phi}{d t}=\mathrm{A} \frac{d B}{d t}=8 \times 2 \times 10^{-4} \times 0.02$ $\varepsilon=3.2 \times 10^{-5} \mathrm{~V}$
Induced current, $\mathrm{I}=\frac{\varepsilon}{R}=2 \times 10^{-5} \mathrm{~A}$
Power loss $=\mathrm{I}^2 \mathrm{R}=4 \times 10^{-10} \times 1.6=6.4 \times 10^{-10} \mathrm{~W}$
Question 3.
A current of 2 A flowing through a coil of 100 turns gives rise to a magnetic flux of $5 \times 10^{-5} \mathrm{~Wb}$ per turn. What is the magnetic energy associated with the coil?
Solution:
Self inductance of coil, $\mathrm{L}=\frac{N \Phi}{I}=\frac{100 \times 5 \times 10^{-3}}{2}$ $=2.5 \times 10^{-3} \mathrm{H}$
Magnetic energy associated with inductance,
$\mathrm{U}=\frac{1}{2} \mathrm{LI}^2=\frac{1}{2} \times 2.5 \times 10^{-3} \times(2)^2$
$=\frac{1}{2} \times 2.5 \times 10^{-3} \times 4=5 \times 10^{-3} \mathrm{~J}$
Question 4.
A transformer is used to light a $140 \mathrm{~W}, 24 \mathrm{~V}$ bulb from a $240 \mathrm{~V}$ AC mains. The current in the main cable is $0.7 \mathrm{~A}$. Find the efficiency of the transformer.
Solution:
$
\begin{aligned}
\text { Efficiency }= & \frac{\text { Output power }}{\text { Input power }} \times 100 \\
\eta= & \frac{140}{240 \times 0.7} \times 100=83.3 \%
\end{aligned}
$

Question 5.
In an ideal step up transformer the turns ratio is $1: 10$. A resistance of $200 \mathrm{ohm}$ connected across the secondary is drawing a current of $0.5 \mathrm{~A}$. What are the primary voltage and current?
Solution:
$
\begin{aligned}
\frac{\mathrm{I}_p}{\mathrm{I}_s} & =\frac{\mathrm{E}_s}{\mathrm{E}_p}=\frac{\mathrm{N}_s}{\mathrm{~N}_p} \\
\mathrm{I}_p & =\frac{\mathrm{N}_s}{\mathrm{~N}_p} \times \mathrm{I}_s=10 \times 0.5=5 \mathrm{~A} \\
\mathrm{E}_s & =\mathrm{I}_s \mathrm{R}_s=0.5 \times 200=100 \mathrm{~V} \\
\mathrm{E}_p & =\frac{\mathrm{E}_s \mathrm{~N}_p}{\mathrm{~N}_s}=\frac{100 \times 1}{10}=10 \mathrm{~V}
\end{aligned}
$
Primary current, $\mathrm{I}_{\mathrm{p}}=5 \mathrm{~A}$
Promary voltage, $\mathrm{E}_{\mathrm{p}}=10 \mathrm{~V}$
Question 6.
A capacitor of capacitance $2 \mu \mathrm{F}$ is connected in a tank circuit oscillating with a frequency of $1 \mathrm{kHz}$. If the current flowing in the circuit is $2 \mathrm{~mA}$, then find the voltage across the capacitor.
Solution:
$
\begin{aligned}
\mathrm{X}_{\mathrm{C}} & =\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi f \mathrm{C}} \\
\varepsilon & =\mathrm{IX}_{\mathrm{C}}=\frac{\mathrm{I}}{2 \pi f \mathrm{C}}=\frac{2 \times 10^{-3}}{2 \times 3.14 \times 1000 \times 2 \times 10^{-6}}=\frac{1}{6.28} \\
\varepsilon & =0.16 \mathrm{~V}
\end{aligned}
$

Question 7.
An ideal inductor takes a current of $10 \mathrm{~A}$ when connected to a $125 \mathrm{~V}, 50 \mathrm{~Hz}$ AC supply. A pure resistor across the same source takes $12.5 \mathrm{~A}$. If the two are connected in series across a $100 \sqrt{2} \mathrm{~V}, 40$ $\mathrm{Hz}$ supply, then calculate the current through the circuit.
Solution:
$
\begin{aligned}
\omega \mathrm{L} & =2 \pi f \mathrm{~L}=\frac{\varepsilon}{\mathrm{I}} \\
2 \pi \mathrm{L} & =\frac{\varepsilon}{f \mathrm{I}}=\frac{125}{50 \times 10}=0.25 \\
\mathrm{R} & =\frac{\varepsilon}{\mathrm{I}}=\frac{125}{12.5}=10 \Omega
\end{aligned}
$
For $40 \mathrm{~Hz}$ frequency,
$
\begin{aligned}
\mathrm{X}_{\mathrm{L}} & =2 \pi \mathrm{L} \times f=2 \pi \mathrm{L} \times 40=10 \Omega \\
\mathrm{Z} & =\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2}=\sqrt{10^2+10^2}=\sqrt{200}=10 \sqrt{2} \Omega \\
\mathrm{I} & =\frac{\varepsilon}{\mathrm{Z}}=\frac{100 \sqrt{2}}{10 \sqrt{2}}=10 \mathrm{~A}
\end{aligned}
$
Question 8.
An LCR series circuit containing a resistance of $120 \Omega$. has angular resonance frequency $4 \times 10^5 \mathrm{rad}$ $\mathrm{s}^{-1}$. At resonance the voltages across resistance and inductance are $60 \mathrm{~V}$ and $40 \mathrm{~V}$, respectively. Find the values of $\mathrm{L}$ and $\mathrm{C}$.

Solution:
$
\begin{aligned}
& \omega=\frac{1}{\sqrt{\mathrm{LC}}} \quad \Rightarrow \mathrm{LC}=\frac{1}{\omega^2}=\frac{1}{16 \times 10^{10}} \\
& \mathrm{I}_{\mathrm{eff}}=\frac{\varepsilon_{\mathrm{R}}}{\mathrm{R}}=\frac{60}{120}=0.5 \mathrm{~A} \\
& \mathrm{I}_{\mathrm{eff}}=\frac{\varepsilon_{\mathrm{L}}}{\omega \mathrm{L}} \quad \Rightarrow \mathrm{L}=\frac{40}{4 \times 10^5 \times 0.5}=20 \times 10^{-5} \mathrm{H} \\
& \mathrm{L}=0.2 \mathrm{mH} \\
& \mathrm{C}=\frac{1}{\omega^2 \mathrm{~L}}=\frac{1}{16 \times 10^{10} \times 0.2 \times 10^{-3}}=\frac{1}{32} \mu \mathrm{F} \\
& \mathrm{C}=\frac{1}{32} \mu \mathrm{F} \\
&
\end{aligned}
$

Question 9.
A coil of inductive reactance $31 \Omega$ has a resistance of $8 \Omega$. It is placed in series with a capacitor of capacitance reactance $25 \Omega$. The combination is connected to an ac source of 110 volt. Find the power factor of the circuit.
Solution:
Power factor $=\frac{R}{Z}=\frac{R}{\sqrt{R^2+\left(X_L-X_C\right)^2}}=\frac{8}{\sqrt{8^2+(31-25)^2}}=\frac{8}{10}$
Power faactor $=0.8$
Question 10.
The power factor of an $\mathrm{RL}$ circuit is $\frac{1}{\sqrt{ } 2}$. If the frequency of $\mathrm{AC}$ is doubled, what will be the power factor?
Solution:
$
\begin{aligned}
\text { Initial power factor } & =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+(\omega \mathrm{L})^2}} \\
& =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{R}^2}}=\frac{\mathrm{R}}{\sqrt{2 \mathrm{R}^2}}=\frac{\mathrm{R}}{\sqrt{2} \mathrm{R}}=\frac{1}{\sqrt{2}}
\end{aligned}
$
$
\begin{aligned}
\text { New power factor } & =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+(2 \omega L)^2}} \\
& =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+4 \mathrm{R}^2}}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}=\frac{1}{\sqrt{5}}
\end{aligned}
$
Question 11.
The instantaneous value of alternating current and voltage are given as $i=\frac{1}{\sqrt{ } 2} \sin (100 \pi \mathrm{t}) \mathrm{A}$ and $\mathrm{e}=$

$\frac{1}{\sqrt{ } 2} \sin \left(100 \pi t+\frac{\pi}{3}\right)$ volt. Find the average power in watts consumed in the circuit.
Solution:
$
\text { Average power } \begin{aligned}
\overline{\mathrm{P}} & =e_{\mathrm{rms}} i_{\mathrm{rms}} \cos \phi \\
& =\frac{e}{\sqrt{2}} \cdot \frac{i}{\sqrt{2}} \cos \phi=\frac{1}{2} \times \frac{1}{2} \times \cos \left(\frac{\pi}{3}\right) \\
\overline{\mathrm{P}} & =\frac{1}{8}
\end{aligned}
$
Common Errors and its Rectifications:
Common Errors:
1. Students sometimes may confuse the peak current and instantaneous value of current and emf.
2. They may confuse in the area of $\mathrm{R}, \mathrm{L}$ and $\mathrm{C}$ with $\mathrm{AC}$. The relation between current and induced emf.
Rectifications:
1. Instantaneous current, $\mathrm{i}=\mathrm{I}_0 \sin$ tot Peak current, $\mathrm{I}_0=\sqrt{ } 2 \mathrm{I}_{\mathrm{rms}}$ Instantaneous emf, $\mathrm{e}=\mathrm{E}_0$ sin cor Peak emf, $\mathrm{E}_0=\sqrt{ } 2 \mathrm{E}_{\mathrm{rms}}$
2. In Inductor: current is $\frac{\pi}{2}$ rad less than that of emf.
In Resistor: current and emf are same phase.
In Capacitor: current is $\frac{\pi}{2}$ rad greater than that of emf.