Numerical Problems

Question 15 Marks

The instantaneous value of alternating current and voltage are given as $i =\frac{1}{\sqrt{2}} \sin (100 \pi t ) A$ and $e=\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}\right)$ volt. Find the average power in watts consumed in the circuit.

Answer

Average power $\overline{ P }=e_{\text {rms }} i_{\text {rms }} \cos \phi$$\begin{aligned}& =\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{ 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 $R, L$ and $C$ with $A C$. The relation between current and induced emf.
Rectifications:
1. Instantaneous current, $i = I _0$ sin tot Peak current, $I _0=\sqrt{ } 2 I _{ rms }$ Instantaneous emf, $e=E_0 \sin$ cor Peak emf, $E_0=\sqrt{ } 2 E_{\text {rms }}$
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.

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Question 25 Marks

The power factor of an RL circuit is $\frac{1}{\sqrt{ } 2}$. If the frequency of AC is doubled, what will be the powe factor?

Answer

$\begin{aligned}\text { Initial power factor } & =\frac{ R }{\sqrt{ R ^2+(\omega L )^2}} \quad[\because \omega L = R ] \\& =\frac{ R }{\sqrt{ R ^2+ R ^2}}=\frac{ R }{\sqrt{2 R ^2}}=\frac{ R }{\sqrt{2} R }=\frac{1}{\sqrt{2}}\end{aligned}$$\begin{aligned}\text { New power factor } & =\frac{R}{\sqrt{R^2+(2 \omega L)^2}} \\& =\frac{R}{\sqrt{R^2+4 R^2}}=\frac{R}{\sqrt{5} R}=\frac{1}{\sqrt{5}}\end{aligned}$

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Question 35 Marks

A coil of inductive reactance $31 \Omega$ has a resistance of $8 \Omega$. It is placed in series with a capacitor 0 capacitance reactance $25 \Omega$. The combination is connected to an ac source of 110 volt. Find the power factor of the circuit.

Answer

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 $=0.8$

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Question 45 Marks

An LCR series circuit containing a resistance of $120 \Omega$. has angular resonance frequency $4 \times 10^5$ rad s$^{-1}$. At resonance the voltages across resistance and inductance are $60 V$ and $40 V$, respectively. Find the values of $L$ and $C$.

Answer

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

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Question 55 Marks

An ideal inductor takes a current of $10 A$ when connected to a $125 V , 50 Hz$ AC supply. A pure resistor across the same source takes $12.5 A$. If the two are connected in series across a $100 \sqrt{ } 2$ $V , 40 Hz$ supply, then calculate the current through the circuit.

Answer

$\begin{aligned}\omega L & =2 \pi / L =\frac{\varepsilon}{ I } \\2 \pi L & =\frac{\varepsilon}{f I }=\frac{125}{50 \times 10}=0.25 \\R & =\frac{\varepsilon}{ I }=\frac{125}{12.5}=10 \Omega\end{aligned}$For $40 Hz$ frequency,$\begin{aligned}X _{ L } & =2 \pi L \times f=2 \pi L \times 40=10 \Omega \\Z & =\sqrt{ R ^2+ X _{ L }^2}=\sqrt{10^2+10^2}=\sqrt{200}=10 \sqrt{2} \Omega \\I & =\frac{\varepsilon}{ Z }=\frac{100 \sqrt{2}}{10 \sqrt{2}}=10 A\end{aligned}$

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Question 65 Marks

A capacitor of capacitance $2 \mu F$ is connected in a tank circuit oscillating with a frequency of 1 $kHz$. If the current flowing in the circuit is $2 mA$, then find the voltage across the capacitor.

Answer

$\begin{aligned}X _{ C } & =\frac{1}{\omega C }=\frac{1}{2 \pi f C } \\\varepsilon & = IX _{ C }=\frac{1}{2 \pi f C }=\frac{1}{2 \times 3.14 \times 1000 \times 2 \times 10^{-6}}=\frac{1}{6.28} \\\varepsilon & =0.16 V\end{aligned}$

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Question 75 Marks

In an ideal step up transformer, the turns ratio is $1: 10$. A resistance of 200 ohm connected across the secondary is drawing a current of $0.5 A$. What are the primary voltage and current?

Answer

$\begin{aligned}\frac{ I _p}{ I _s} & =\frac{ E _s}{ E _p}=\frac{ N _s}{ N _p} \\I _p & =\frac{ N _s}{ N _p} \times I _s=10 \times 0.5=5 A \\E _s & = I _s R _s=0.5 \times 200=100 V \\E _p & =\frac{ E _s N _p}{ N _s}=\frac{100 \times 1}{10}=10 V\end{aligned}$Primary current, $I _{ p }=5 A$

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Question 85 Marks

A transformer is used to light a $140 W , 24 V$ bulb from a $240 V$ AC mains. The current in the main cable is $0.7 A$. Find the efficiency of the transformer.

Answer

$\text { Efficiency }=\frac{\text { Output power }}{\text { Input power }} \times 100 $
$\eta=\frac{140}{240 \times 0.7} \times 100=83.3 \%$

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Question 95 Marks

A current of $2 A$ flowing through a coil of $100$ turns gives rise to a magnetic flux of $5 \times 10^{-5} Wb$ per turn. What is the magnetic energy associated with the coil?

Answer

Self inductance of coil, $L =\frac{N \Phi}{I}=\frac{100 \times 5 \times 10^{-3}}{2}$ $=2.5 \times 10^{-3} H$
Magnetic energy associated with inductance,
$U=\frac{1}{2} L^2=\frac{1}{2} \times 2.5 \times 10^{-3} \times(2)^2 $
$=\frac{1}{2} \times 2.5 \times 10^{-3} \times 4=5 \times 10^{-3} J$

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Question 105 Marks

A rectangular loop of sides $8 cm$ and $2 cm$ is lying in a uniform magnetic field of magnitude 0.5 $T$ with its plane normal to the field. The field is now gradually reduced at the rate of $0.02 T / s$. If the resistance of the loop is $1.6 \Omega$, then find the power dissipated by the loop as heat.

Answer

Induced emf, $|\varepsilon|=\frac{d \Phi}{d t}=A \frac{d B}{d t}=8 \times 2 \times 10^{-4} \times 0.02$

$\varepsilon=3.2 \times 10^{-5} V$

Induced current, $\mid=\frac{\varepsilon}{R}=2 \times 10^{-5} A$

Power loss $=I^2 R=4 \times 10^{-10} \times 1.6=6.4 \times 10^{-10} W$

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Question 115 Marks

A coil has $2000$ turns and area $70 cm ^2$. The magnetic field perpendicular to the plane of the coil is $0.3 Wb / m ^2$. The coil takes $0.1 s$ to rotate through $180^{\circ}$. Then what is the value of induced emf?

Answer

Magnitude of change in flux,
$|\Delta \Phi|=\mid NBA \left(\cos 180^{\circ}-\cos 0^{\circ}\right. $
$=| NBA (-1-1)|=|-2 NBA |=|2 NBA |$
Where,$N =2000 \\B =0.3 Wb / m ^2 $
$A =70 \times 10^{-4} m ^2 $
$t =0.1 sec $
$\text { 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} $
$\varepsilon=84\ V$

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