Alternating Current Test

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Alternating Current Test - 42

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Question 1
1 / -0

In $$LCR$$ series $$AC$$ circuit

A
If $$R$$ is increased current will decrease.

B
If $$L$$ is increased current will decrease.

C
If $$C$$ is increased current will increase.

D
If $$C$$ is increased current will decrease.

Solution

$$\displaystyle I=\frac{V}{Z}$$$$\displaystyle =\frac{V}{\sqrt{R^2+\left(\omega L\sim \frac{1}{\omega C}\right)^2}}$$

By increasing $$R$$, current will definitely decrease by change in $$L$$ or $$C$$, current may increase or decrease.
Question 2
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For an $$LCR$$ series circuit with an A.C. source of angular frequency $$\omega$$, which statement is correct?

A
Circuit will be capacitive if $$\omega > \displaystyle\frac{1}{\sqrt{LC}}$$

B
Circuit will be capacitive if $$\omega = \displaystyle\frac{1}{\sqrt{LC}}$$

C
Power factor of circuit will be unity if capacitive reactance equals inductive reactance

D
Current will be leading voltage if $$\omega > \displaystyle\frac{1}{\sqrt{LC}}$$

Solution

Circuit will be capacitive if , total reactance =$$ X_L - X_C < 0$$
$$ \Rightarrow X_C > X_L \Rightarrow \frac{1}{C \omega} > L \omega \Rightarrow \omega < \frac{1}{ \sqrt{ L C }} $$

Current will be leading the voltage if the circuit is capacitive , i.e., $$ \omega < \frac{1}{ \sqrt{ L C }} $$

Power factor= $$ \dfrac{ R } { Z } $$ ; pf factor will be unity if $$ R= Z \Rightarrow X_L=X_C $$ , i.e., capacitive reactance is equal to inductive reactance.
Question 3
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A $$220V$$, $$50\space Hz$$ AC generator is connected to an inductor and a $$50\space \Omega$$ resistance in series. The current in the circuit is $$1.0\space A$$. What is the Potential difference across inductor?

A
$$102.2\space V$$

B
$$186.4\space V$$

C
$$214\space V$$

D
$$170\space V$$

Solution

$$\left | Z \right| = \dfrac{V_{rms}}{I_{rms} } = \dfrac{220}{1} =220 \Omega $$

$$ X_L = \sqrt{ {\left | Z \right | }^2 - R^2 } = 214 H $$

$$ V_L = \dfrac{ X_L}{ \left | Z \right | } V_{rms} = \dfrac{ 214}{220} 220 = 214 V $$
Question 4
1 / -0

An alternating voltage $$E = 50\sqrt2\sin(100t)\space V$$ is connected to a $$1\space \mu F$$ capacitor through an ac ammeter. What will be the reading of the ammeter?

A
$$10\space mA$$

B
$$5\space mA$$

C
$$5\sqrt2\space mA$$

D
$$10\sqrt2\space mA$$

Solution

$$ X_C = \frac{1}{C \omega} = 10000 \Omega $$
ammeater reading = $$ I_{rms} = \frac{ V_{rms} }{ \left | jX_C \right | } = \frac{ 50 }{10000} = 5 mA $$
Question 5
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A $$50\space W$$, $$100\space V$$ lamp is to be connected to an AC mains of $$200\space V, \space 50\space Hz$$. What capacitor is essential to be put in series with the lamp?

A
$$\displaystyle\frac{25}{\sqrt2}\mu F$$

B
$$\displaystyle\frac{50}{\pi\sqrt3}\mu F$$

C
$$\displaystyle\frac{50}{\sqrt2}\mu F$$

D
$$\displaystyle\frac{100}{\pi\sqrt3}\mu F$$

Solution

for $$100V, 50 W $$lamp , $$ R = \frac{100^2}{50} = 200 \Omega $$
capacitor put in series should be such that,
$$ V_R = \frac{ R} { \sqrt{ R^2 + X_C^2 } } V_s= 100 $$
$$ \Rightarrow \frac{1}{C \omega} = X_C= \sqrt{3} R $$
$$ C=\frac{1}{ \omega \sqrt{3} R } =\frac{50}{\pi \sqrt{3} } \mu F $$
Question 6
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A current source sends a current $$I - i_0\cos(\omega t)$$. when connected across an unknown load, it gives a voltage output of $$v = v_0\sin[\omega t + (\pi/4)]$$ across that load. then the voltage across the current source may be brought in phase with the current through it by

A
Connecting an inductor in series with the load

B
Connecting a capacitor in series with the load

C
Connecting an inductor in parallel with the load

D
Connecting a capacitor in parallel with the load

Solution

$$ v = v_0\sin[\omega t + (\pi/4)] = v_0\sin[\omega t + (\pi/4) + (\pi/4) - (\pi/4)] = v_0\sin[\omega t +(\pi/2) - (\pi/4)] = v_0\cos[\omega t - (\pi/4)] $$
cuurent = $$ I - i_0 \cos[\omega t ] = I + i_0 \cos[\omega t + \pi ] $$
we therefore see that ac current leads the ac voltage across the load. if an inductor is connected in series with the load, the it induces a lag in the current wrt to the voltage.
Therefore an inductor must be connected in series with the load so as to bring the voltage across the current source in phase with the current.
Question 7
1 / -0

Current in an ac circuit is given by $$I = 3\sin\omega t + 4\cos\omega t$$, then

A
RMS value of current is $$5\space A$$

B
Mean value of this current in any one half period will be $$6\pi$$

C
If voltage applied is $$V = V_m\sin\omega t$$, then the circuit may be containing resistance and capacitance only

D
If voltage applied is $$V = V_m\sin\omega t$$, the circuit may contain resistance and inductance only

Solution

$$ I = 3sin \omega t + 4 cos \omega t $$
$$\Rightarrow I= 5( \frac{3}{5} sin \omega t + \frac{4}{5} cos \omega t )$$
$$ \Rightarrow I = 5 sin( \omega t + \alpha ) $$ $$ ; \alpha = tan^{-1} {\frac{4}{5} } $$
therefore,
rms value of current = $$ 5/ \sqrt{2} A $$
Mean value of current one half period = $$ \frac { 1 }{ \pi } \int _{ 0 }^{ \pi }{ 5sin\theta \quad d\theta } =\frac { 1 }{ \pi } 5\quad { \left| -cos\theta \right| }_{ 0 }^{ \pi }=\frac { 10 }{ \pi } \quad A $$
The current lags the voltage by an angle $$ \theta $$, therefore the circuit may contain resistance and capacitor only.
Question 8
1 / -0

In an ac circuit, the potential differences across an inductance and resistance joined in series are, respectively, $$16\space V$$ and $$20\space V$$. The total potential difference across the circuit is

A
$$20\space V$$

B
$$25.6\space V$$

C
$$31.9\space V$$

D
$$53.5\space V$$

Solution

In phasor, $$V_R= 20 \angle{0} $$; $$ V_L= 16 \angle{(-90)} $$
Total potential difference is = $$ V_R+V_L= 25.6 \angle{(-38.66)} $$
Magnitude of total potential difference =$$25.6 V$$
Question 9
1 / -0

Voltage across each elements of a series LCR circuit are given by $$V_L= 60V, V_C = 20V, V_R= 30V$$ Find out source voltage

A
50V

B
100V

C
150V

D
200V

Solution

$$V=\sqrt { { { V }_{ R } }^{ 2 }+{ \left( { V }_{ L }-{ V }_{ C } \right) }^{ 2 } } =\sqrt { { 30 }^{ 2 }+{ \left( 60-20 \right) }^{ 2 } } =50V$$
Question 10
1 / -0

What is the value of inductance L for which the current is a maximum in a series LCR circuit with $$C=10 \mu F $$ and $$\omega = 1000s^{-1}$$ ?

A
$$10 mH$$

B
$$100mH$$

C
$$1 mH$$

D
cannot be calculated unless R is known

Solution

Current is maximum at resonance

$$\displaystyle \Rightarrow \omega^2 = \frac{1}{LC} \Rightarrow L = \frac{1}{\omega^2 C}$$

$$\displaystyle = \frac{1}{(1000)^2 (10 \times 10^{-6})} = 0.1 H =100 mH$$

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 42

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Alternating Current Test - 42

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Question 1

If $$R$$ is increased current will decrease.

If $$L$$ is increased current will decrease.

If $$C$$ is increased current will increase.

If $$C$$ is increased current will decrease.

Solution

Question 2

Circuit will be capacitive if $$\omega > \displaystyle\frac{1}{\sqrt{LC}}$$

Circuit will be capacitive if $$\omega = \displaystyle\frac{1}{\sqrt{LC}}$$

Power factor of circuit will be unity if capacitive reactance equals inductive reactance

Current will be leading voltage if $$\omega > \displaystyle\frac{1}{\sqrt{LC}}$$

Solution

Question 3

$$102.2\space V$$

$$186.4\space V$$

$$214\space V$$

$$170\space V$$

Solution

Question 4

$$10\space mA$$

$$5\space mA$$

$$5\sqrt2\space mA$$

$$10\sqrt2\space mA$$

Solution

Question 5

$$\displaystyle\frac{25}{\sqrt2}\mu F$$

$$\displaystyle\frac{50}{\pi\sqrt3}\mu F$$

$$\displaystyle\frac{50}{\sqrt2}\mu F$$

$$\displaystyle\frac{100}{\pi\sqrt3}\mu F$$

Solution

Question 6

Connecting an inductor in series with the load

Connecting a capacitor in series with the load

Connecting an inductor in parallel with the load

Connecting a capacitor in parallel with the load

Solution

Question 7

RMS value of current is $$5\space A$$

Mean value of this current in any one half period will be $$6\pi$$

If voltage applied is $$V = V_m\sin\omega t$$, then the circuit may be containing resistance and capacitance only

If voltage applied is $$V = V_m\sin\omega t$$, the circuit may contain resistance and inductance only

Solution

Question 8

$$20\space V$$

$$25.6\space V$$

$$31.9\space V$$

$$53.5\space V$$

Solution

Question 9

50V

100V

150V

200V

Solution

Question 10

$$10 mH$$

$$100mH$$

$$1 mH$$

cannot be calculated unless R is known

Solution

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