Alternating Current Test

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Alternating Current Test - 75

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Question 1
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A $$200$$km long telegraph wire has a capacitance of $$0.014$$ $$\mu F/km$$. If it carries an alternating current of $$50\times 10^3$$Hz, what should be the value of an inductance required to be connected in series so that impedance is minimum?

A
$$0.48\times 10^{-2}$$mH

B
$$0.36\times 10^{-2}$$mH

C
$$0.52\times 10^{-2}$$mH

D
$$0.49\times 10^{-2}$$mH

Solution

Given, $$C'=0.014\mu F/km$$
$$f=50\times 10^3$$Hz
Total capacitance,
$$C=C'\times 200$$
$$=0.014\mu F\times 200=2.8\mu F$$
For impedance to be minimum
$$X_L=X_C,$$ i.e., $$\omega L=\displaystyle\frac{1}{\omega C}$$
$$\displaystyle L=\frac{1}{\omega ^2C}$$
$$=\displaystyle\frac{1}{4\pi^2l^2C}$$
$$=\displaystyle\frac{1}{4\pi^2\times (50\times 10^3)^2\times 2.8\times 10^{-6}}$$
$$=0.36\times 10^{-3}H=0.36\times 10^{-2}$$mH.
Question 2
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In an L-R circuit, the voltage is given by $$V$$ as $$283 sin 314t$$. The current is found tc be $$4 \, sin \left ( 314t-\dfrac {\pi}{4} \right )$$Calculate the resistance of the circuit.

A
$$25 \Omega$$

B
$$50\Omega$$

C
$$75\Omega$$

D
$$90\Omega$$

Solution

In L-R series circuit current lags the voltage by an angle,
$$\phi =tan^{-1}\left ( \dfrac{X_L}{R} \right )$$
$$\phi =\dfrac{\pi}{4}, X_L=R, \omega L=R$$
Given, $$V= 283 \, sin \, 314t$$
Comparing with standard equation
$$V=V_0 sin\, \omega\, t$$
$$\omega =314$$
$$314 L =R$$
Further $$V_0 =i_0 |Z|$$
$$283 = 4\sqrt{R^2+X^2_L}$$
$$R^2 +(\omega L)^2=\left ( \dfrac{283}{4} \right )^2=5005.86$$
$$2R^2 = 5005.56$$
$$R= 50 \Omega $$
Question 3
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A RC series circuit of R$$=15\Omega$$ and $$C=10\mu F$$ is connected to $$20$$ volt DC supply for very long time. Then capacitor is disconnected from circuit and connected to inductor of $$10$$mH. Find amplitude of current.

A
$$0.2\sqrt{10}$$A

B
$$2\sqrt{10}$$A

C
$$0.2$$A

D
$$\sqrt{10}$$A

Solution

Given : $$C = 10\mu F$$ $$V_o = 20$$ volts $$L = 10mH$$
Amplitude of the current $$I_o=Q_o w=(CV_o)\dfrac{1}{\sqrt{LC}} = \sqrt{\dfrac{C}{L}}V_o$$
$$\implies \ $$ $$I_o = \sqrt{\dfrac{10\times 10^{-60}}{10\times 10^{-3}}}\times 20 = 0.2\sqrt{10} A$$
Question 4
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An L-C-R series circuit containing a resistance of $$R = 120 \Omega $$ has angular resonant frequency $$4 \times 10^5$$ rad/s. At resonance the voltage across resistance and inductance are 60 V and 90 V respectively. Then values of L and C are :

A
$$0.2 \, mH, \dfrac{1}{16}\mu F$$

B
$$0.2 \, mH, \dfrac{1}{32}\mu F$$

C
$$0.4 \, mH, \dfrac{1}{32}\mu F$$

D
$$0.4 \, mH, \dfrac{1}{16}\mu F$$

Solution

$$R= 120 \Omega , \omega _r=4 \times 10^5 $$ rad/sec
$$V_R=60V, V_L=40V$$
$$V_R=iR \Rightarrow 60 = i \times 120$$
$$i= \dfrac{1}{2}A$$
$$V_L=iX_L\Rightarrow 40=\dfrac{1}{2}X_L$$
$$\therefore X_L=80 \Omega $$
$$\omega _rL=80\Rightarrow 4 \times 10^5 L=80$$
$$L=20 \times 10^{-5}H=0.2 mH$$
$$X_C=X_L=80 \Rightarrow \dfrac{1}{\omega _rC}=80$$
$$C=\dfrac{1}{80\omega }=\dfrac{1}{80\times 4\times 10^5}$$
$$\dfrac{10^{-6}}{32}=\dfrac{1}{32}\mu F$$
Question 5
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In the circuit diagram shown, $$X_C=100 \Omega,X_L=200 \Omega\, and \, R=100 \Omega$$. Effective current through the source is :

A
$$2A$$

B
$$2\sqrt{2}A$$

C
$$0.5 A$$

D
$$\sqrt {0.4} A$$

Solution
Question 6
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Two parallel wires in the plane of the paper are distance $$X_0$$ apart. A point charge is moving with speed u between the wires in the same plane at a distance $$X_1$$ from one of the wires. When the wires carry current of magnitude I in the same direction, the radius of curvature of the path of the point charge is $$R_1$$. In contract, if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is $$R_2$$. If $$\dfrac{X_0}{X_1}=3$$, the value of $$\dfrac{R_1}{R_2}$$ is?

A
6

B
$$3$$.

C
4

D
5

Solution
Question 7
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The equivalent inductance between $$A$$ and $$B$$ is:

A
$$1 H$$

B
$$4 H$$

C
$$0.8 H$$

D
$$16 H$$

Solution

Here, all the inductances are connected in parallel.
Hence, the equivalent inductance between A and B is
$$\displaystyle \frac{1}{L_{AB}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4} = 1$$
or $$L_{AB} = 1 H$$
Question 8
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An alternating voltage given as $$V=100\sqrt { 2 } \sin { 100t } \quad $$is applied to a capacitor of $$1\mu F$$. The current reading of the ammeter will be equal to ______ mA.

A
$$10$$

B
$$20$$

C
$$40$$

D
$$80$$

Solution

Given : $$C = 1\mu F$$
On comparing the given voltage with $$V = V_o \ \sin wt$$
we get $$V_o =100\sqrt{2}$$ and $$w = 100$$
Capacitive reactance $$X_C = \dfrac{1}{wC} = \dfrac{1}{100\times 10^{-6}} = 10^4\Omega$$
Rms value of voltage $$V_{rms} = \dfrac{V_o}{\sqrt{2}} = 100$$ volts
Thus reading of ammeter $$I_{rms} = \dfrac{V_{rms}}{I_{rms}} = \dfrac{100}{10^4} = 0.01 \ A = 10 \ mA$$
Question 9
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An electrical device draws $$2 kW$$ power from ac mains voltage $$223 V(rms).$$ The current differs lags in phase by $$\phi = tan^{-1} \left ( -\dfrac{3}{4} \right )$$ as compared to voltage. The resistance R in the circuit is:

A
$$15\Omega$$

B
$$20 \Omega$$

C
$$25 \Omega$$

D
$$30 \Omega$$

Solution

Here, $$P \, = \, 2 \, kW \, = \, 2 \, \times \, 10^3W$$
$$V_{rms} \, = \, 233 \, V, tan \, \phi \, = \, -\dfrac{3}{4}$$
$$As, \, P \, = \, \dfrac{V^2_{rms}}{Z}$$

$$\Rightarrow \, Z \, = \, \dfrac{V^2_{rms}}{P} \, = \, \dfrac{(223)^2}{2000} \, = \dfrac{49729}{2000} \, = \, 24.86 \, \Omega \, or \, Z \, = \, 25 \, \Omega$$

$$\tan \, \phi \, = \, \dfrac{X_C \, - \, X_L}{R} \, = \, - \dfrac{3}{4} \, \therefore \, X_C \, - \, X_L \, = \, -\dfrac{3}{4} R.$$

AS, $$Z^2 \, = \, R^2 \, + \, (X_C \, - \, X_L)^2$$

$$\therefore \, (25)^2 \, = \, R^2 \, + \, \left(-\dfrac{3}{4} \, R \right)^2$$

$$625 \, = \, \dfrac{25 \, R^2}{16}.$$

$$R^2 \, = \, \dfrac{625 \, \times \, 16}{25} \, \, \Rightarrow \, R \, = \, 20 \, \Omega$$
Question 10
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In an oscillating $$LC$$ circuit the maximum charge on the capacitor is $$Q$$. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is

A
$$Q/2$$

B
$$Q/\surd 3$$

C
$$Q/\surd 2$$

D
$$Q$$

Solution

[C] When capacitor is full charge, total energy of LC circuit is with capacitor
$$\begin{array}{l} E=\dfrac { 1 }{ 2 } \, \, \dfrac { { { Q^{ 2 } } } }{ C } ,Q \to max\, \, { { charge } }\, \, on\, \, capacitor \\ max\, \, energy\, \, in\, \, Inductor=\dfrac { { LI_{ 0 }^{ 2 } } }{ 2 } ,{ I_{ 0 } }=current \end{array}$$
According to the question
Let q charge on capacitor when charge in equally shared between inductor and capacitor.
$$\begin{array}{l} \therefore \dfrac { 1 }{ 2 } \left( { \dfrac { 1 }{ 2 } \dfrac { { { Q ^{ 2 } } } }{ C } } \right) =\dfrac { 1 }{ 2 } \dfrac { { { q^{ 2 } } } }{ C } \\ \Rightarrow { Q^{ 2 } }=2{ q^{ 2 } } \\ \therefore q=\dfrac { Q}{ { \sqrt { 2 } } } \end{array}$$