Alternating Current Test

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Alternating Current Test - 73

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Question 1
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A capacitor of $$10\space \mu F$$ and an inductor of $$1\space H$$ are joined in series. An ac of $$50\space Hz$$ is applied to this combination. What is the impedance of the combination?

A
$$\displaystyle\frac{5(\pi^2 - 5)}{\pi}\space \Omega$$

B
$$\displaystyle\frac{10(10 - \pi^2)}{\pi}\space \Omega$$

C
$$\displaystyle\frac{10(\pi^2 - 5)}{\pi}\space \Omega$$

D
$$\displaystyle\frac{5(10 - \pi^2)}{\pi}\space \Omega$$

Solution

$$ X_L = L \omega = L 2 \pi f = 100 \pi \Omega $$
$$ X_C = \frac{ 1 } { C \omega } = \frac{ 1 } { C 2 \pi f } = 1000/ \pi \Omega $$
impedance =$$ \left | X_L - X_C \right | = 100 \dfrac{ 10- {\pi}^2 }{\pi} \Omega $$
Question 2
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A transmitter transmits at a wavelength of $$300\space m$$. A condenser of capacitance $$2.4\space \mu F$$ is being used. The value of the inductance for the resonant circuit is approximately

A
$$10^{-4}\space H$$

B
$$10^{-6}\space H$$

C
$$10^{-8}\space H$$

D
$$10^{-10}\space H$$

Solution

$$ f = \dfrac{c}{ \lambda} = 3 \times 10^8 / 300 $$

$$ \omega = 2\pi f = \dfrac{1}{\sqrt{L C }} $$

$$ L = \dfrac{ 1}{C 4 {\pi}^{2} f^2 }= 10^{-8} H $$
Question 3
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Directions For Questions

It is known to all of you that the impedance of a circuit is dependent on the frequency of source. In order to study the effect of frequency on the impedance, a student in a lab took $$2$$ impedance boxes $$P$$ and $$Q$$ and connected them in series with an $$AC$$ source of variable frequency. The emf of the source is constant at $$10V$$. Box $$P$$ contains a capacitance of $$1\mu F$$ in series with a resistance of $$32\Omega$$. And the box $$Q$$ has a coil of self inductance $$4.9 mH$$ and a resistance of $$68\Omega$$ in series. He adjusted the frequency so that the maximum current flows in $$P$$ and $$Q$$. Based on his experimental set up and the reading by him at various moment answer the following questions.

...view full instructions

Impedance of box $$P$$ at the above frequency is

A
$$70\Omega$$

B
$$77\Omega$$

C
$$90\Omega$$

D
$$100\Omega$$

Solution

$$\displaystyle X_C=\frac{1}{\omega C}=\frac{1}{\left(\displaystyle \frac{10^5}{7}\right)(10^{-6})}=70\Omega$$
$$\displaystyle Z_P=\sqrt{R^2_P+X^2_C}$$
$$=\displaystyle \sqrt{(32)^2+(70)^2}=77\Omega$$
Question 4
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Directions For Questions

It is known to all of you that the impedance of a circuit is dependent on the frequency of source. In order to study the effect of frequency on the impedance, a student in a lab took $$2$$ impedance boxes $$P$$ and $$Q$$ and connected them in series with an $$AC$$ source of variable frequency. The emf of the source is constant at $$10V$$. Box $$P$$ contains a capacitance of $$1\mu F$$ in series with a resistance of $$32\Omega$$. And the box $$Q$$ has a coil of self inductance $$4.9 mH$$ and a resistance of $$68\Omega$$ in series. He adjusted the frequency so that the maximum current flows in $$P$$ and $$Q$$. Based on his experimental set up and the reading by him at various moment answer the following questions.

...view full instructions

The angular frequency for which he detects maximum current in the circuit is

A
$$10^5/7 rad/s$$

B
$$10^4 rad/s$$

C
$$10^5 rad/s$$

D
$$10^4/7 rad/s$$

Solution

$$\displaystyle \omega =\frac{1}{\sqrt {LC}}$$
$$=\displaystyle \frac{1}{\sqrt{4.9\times 10^{-3}\times 10^{-6}}}$$
$$\displaystyle =\frac{10^5}{7} rad/s$$
Question 5
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A resistor and a capacitor are connected to an A.C. supply of $$200\space V, \space 50\space Hz$$ in series. The current in the circuit is $$2\space A$$. If the power consumed in the circuit is $$100\space W$$ then the capacitive reactance in the circuit is

A
$$100\space \Omega$$

B
$$25\space \Omega$$

C
$$\sqrt{125\times75}\space \Omega$$

D
$$400\space \Omega$$

Solution

$$ V_{rms} = 200 V $$
$$I_{rms} = 2A $$
$$ P = 100 = I_{rms}^2 R = 4 R $$
$$ R=25 \Omega $$

$$\therefore X_C= \sqrt { R^2 + X_C^2 } = \dfrac {V_{rms}}{I_{rms}} =100 $$

$$ \Rightarrow X_C= \sqrt{ 100^2 - 25^2 } = \sqrt{ 9375} = \sqrt{ 125 \times 75} \Omega$$
Question 6
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A $$50\ Hz$$ a.c. source of $$20\ V$$ is connected across R and C as shown in figure below. The voltage across R is $$12\ V$$. The voltage across C is:

A
$$8\ V$$

B
$$16\ V$$

C
$$10\ V$$

D
not possible to determine unless values of R and C are given

Solution

For ac C- R circuit,
$$V_S^2=V_R^2+V_C^2$$
or
$$V_C=\sqrt{V_S^2-V_R^2}$$

$$=\sqrt{20^2-12^2}$$

$$=\sqrt{400-144}$$
$$=\sqrt{256}=16\ V$$
Question 7
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A series $$LCR$$ circuit containing a resistance of $$120\space \Omega$$ has angular resonance frequency $$4\times10^5\space rad\space s^{-1}$$. At resonance the voltages across resistance and inductance are $$60\space V$$ and $$40\space V$$, respectively. The value of inductance $$L$$ is

A
$$0.1\space mH$$

B
$$0.2\space mH$$

C
$$0.35\space mH$$

D
$$0.4\space mH$$

Solution

At resonance the reactance of inductor and the capaciotr cancel each other,
$$L\omega = \frac{1}{C\omega} $$ and total $$Z= R \Omega $$
resonance frequency = $$ {\omega}_r = 4 \times 10^5 rad/s $$
let the current at resonance be $$I$$.
$$V_R = I \times R = 60 V \Rightarrow I= \frac{60}{120} A $$
$$ V_L = I \times L{\omega}_r = 40 V \Rightarrow L=0.2 mH $$
Question 8
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Directions For Questions

A box $$P$$ and a coil $$Q$$ are connected in series with an ac source of variable frequency. The emf of the source is constant at $$10\space V$$. Box $$P$$ contains a capacitance of $$1\space \mu F$$ in series with a resistance of $$32\space \Omega$$. Coil $$Q$$ has a self inductance of $$4.9\space mH$$ and a resistance of $$68\space \Omega$$ in series. The frequency is adjusted so that maximum current flows in $$P$$ and $$Q$$.

...view full instructions

The impedance of $$P$$ at this frequency is :

A
$$77\space \Omega$$

B
$$36\space \Omega$$

C
$$40\space \Omega$$

D
$$125\space \Omega$$

Solution

P has $$ R_1= 32 \Omega $$ $$ C_1=1 \mu F $$
Q has $$ R_2=68 \Omega $$ $$ L_2 = 4.8 m H $$
Total impedance of the series combination of P and Q :
$$ Z= R_1+R_2 +j L\omega -j\frac{1}{C\omega} $$
The frequency is adjusted so that maximum current flows in P and Q. that is, frequency is adjusted to have minimum total series impedance. Total series impedance will be minimum when the reactance of L and C cancel each other. Therefore,
$$ \omega = \frac{1}{\sqrt{LC}} =\frac{100000}{7} rad/s$$
Therefore at maximum current, impedance of P= $$ R_1 + -j\frac{1}{C\omega} = 32 -70j =77 \Omega $$
Question 9
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Directions For Questions

A current of $$4\space A$$ flows in a coil when connected to a $$12\space V$$ dc source. If the same coil is connected to a $$12\space V$$, $$50\space rad\space s^{-1}$$ ac source, a current of $$2.4\space A$$ flows in the circuit.

...view full instructions

The inductance of the coil is

A
$$0.02\space H$$

B
$$0.04\space H$$

C
$$0.08\space H$$

D
$$1.0\space H$$

Solution

impedance of coil=$$ R +jL\omega$$.
when DC is applied,
$$ I_{dc}= \frac{V_{dc}}{R} \Rightarrow 4 = \frac{12}{R} \Rightarrow R= 3\Omega $$
when AC(12 V , 50rad/s ) is applied,
$$ I_{ac} = \frac{ V_{ac}}{ \sqrt{ R^2+{(L \omega )}^2 } } \Rightarrow L = \sqrt{ \frac{1}{{\omega}^2}(\frac{V_{ac}^2}{I_{ac}^2} - R^2) } = \frac{2}{25} = .08 H $$
Question 10
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Directions For Questions

A box $$P$$ and a coil $$Q$$ are connected in series with an ac source of variable frequency. The emf of the source is constant at $$10\space V$$. Box $$P$$ contains a capacitance of $$1\space \mu F$$ in series with a resistance of $$32\space \Omega$$. Coil $$Q$$ has a self inductance of $$4.9\space mH$$ and a resistance of $$68\space \Omega$$ in series. The frequency is adjusted so that maximum current flows in $$P$$ and $$Q$$.

...view full instructions

The impedance of $$Q$$ at this frequency is:

A
$$200\space \Omega$$

B
$$\sqrt{1350}\space \Omega$$

C
$$55\space \Omega$$

D
$$\sqrt{9524}\space \Omega$$

Solution

P has $$ R_1= 32 \Omega $$ $$ C_1=1 \mu F $$
Q has $$ R_2=68 \Omega $$ $$ L_2 = 4.9 m H $$
Total impedance of the series combination of P and Q :
$$ Z= R_1+R_2 +j L\omega -j\frac{1}{C\omega} $$
The frequency is adjusted so that maximum current flows in P and Q. that is, frequency is adjusted to have minimum total series impedance. Total series impedance will be minimum when the reactance of L and C cancel each other. Therefore,
$$ \omega = \frac{1}{\sqrt{LC}} =\frac{100000}{7} rad/s$$
Therefore at maximum current, impedance of $$ Q= R_2 + jL\omega = 68+70j =\sqrt{9524} \Omega $$