Alternating Current MCQ Questions & Answers in Electrostatics and Magnetism | Physics
Learn Alternating Current MCQ questions & answers in Electrostatics and Magnetism are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
21.
A series $$LR$$ circuit is connected to an $$ac$$ source of frequency $$\omega $$ and the inductive reactance is equal to $$2R.$$ A capacitance of capacitive reactance equal to $$R$$ is added in series with $$L$$ and $$R.$$ The ratio of the new power factor to the old one is
22.
In the given circuit, the reading of voltmeter $${V_1}$$ and $${V_2}$$ are $$300\,V$$ each. The reading to the voltmeter $${V_3}$$ and ammeter $$A$$ are respectively
A
$$150\,V,2.2\,A$$
B
$$220\,V,2.2\,A$$
C
$$220\,V,2.0\,A$$
D
$$100\,V,2.0\,A$$
Answer :
$$220\,V,2.2\,A$$
For series $$LCR$$ circuit
Voltage $$V = \sqrt {V_R^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} $$
Since, $${V_L} = {V_C}$$
Hence $$V = {V_R} = 220\,V$$
Also, current $$i = \frac{V}{R} = \frac{{220}}{{100}} = 2.2\,A$$
23.
Power dissipated in an $$L-C-R$$ series circuit connected to an $$AC$$ source of emf $$\varepsilon $$ is
Power dissipated in series $$LCR.$$
$$\eqalign{
& P = I_{rms}^2R \cr
& = \frac{{\varepsilon _{rms}^2R}}{{{{\left| Z \right|}^2}}} = \frac{{{\varepsilon ^2}R}}{{\left[ {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} \right]}} \cr} $$
As $${I_{rms}} = \frac{{{V_{rms}}}}{Z}$$
24.
In the $$LC$$ circuit, the current is in the direction shown and the charges on the capacitor plates have the signs shown. At this time
A
$$I$$ is increasing and $$Q$$ is increasing
B
$$I$$ is increasing and $$Q$$ is decreasing
C
$$I$$ is decreasing and $$Q$$ is increasing
D
$$I$$ is decreasing and $$Q$$ is decreasing
Answer :
$$I$$ is increasing and $$Q$$ is decreasing
As it can be easily seen by the direction of $$I$$ that $$Q$$ is decreasing thus, energy of capacitor is decreasing and hence, energy of inductance is increasing or $$\left( {\frac{1}{2}L{I^2}} \right)$$ gives that $$I$$ is increasing.
25.
A series $$R-C$$ circuit is connected to an alternating voltage source. Consider two situations :
1. When capacitor is air filled.
2. When capacitor is mica filled.
Current through resistor is $$i$$ and voltage across capacitor is $$V$$ then
A
$${V_a} < {V_b}$$
B
$${V_a} > {V_b}$$
C
$${i_a} > {i_b}$$
D
$${V_a} = {V_b}$$
Answer :
$${V_a} > {V_b}$$
Net reactive capacitance,
$${X_C} = \frac{1}{{2\pi fC}}$$
So, current in circuit, $$I = \frac{V}{Z} = \frac{V}{{\sqrt {{R^2} + {{\left( {\frac{1}{{2\pi fC}}} \right)}^2}} }}$$
$$ \Rightarrow I = \frac{{2\pi fC}}{{\sqrt {4{\pi ^2}{f^2}{C^2}{R^2} + 1} }} \times V$$
Voltage drop across capacitor, $${V_C} = I \times {X_C}$$
$$ = \frac{{2\pi fC}}{{\sqrt {4{\pi ^2}{f^2}{C^2}{R^2} + 1} }} \times \frac{1}{{2\pi fC}}$$
i.e. $${V_C} = \frac{V}{{\sqrt {4{\pi ^2}{f^2}{C^2}{R^2} + 1} }}$$
When mica is introduced, capacitance will increase hence, voltage across capacitor get decrease.
26.
When the $$rms$$ voltages $${V_L},{V_C}$$ and $${V_R}$$ are measured respectively across the inductor $$L,$$ the capacitor $$C$$ and the resistor $$R$$ in a series $$LCR$$ circuit connected to an $$AC$$ source, it is found that the ratio $${V_L}:{V_C}:{V_R} = 1:2:3.$$ If the $$rms$$ voltage of the $$AC$$ sources is $$100\,V,$$ the $${V_R}$$ is close to :
A
$$50\,V$$
B
$$70\,V$$
C
$$90\,V$$
D
$$100\,V$$
Answer :
$$90\,V$$
$$\eqalign{
& {\text{Given,}}\,{V_L}:{V_C}:{V_R} = 1:2:3 \cr
& V = 100\,V \cr
& {V_R} = ? \cr} $$
As we know, $$V = \sqrt {V_R^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} $$
Solving we get, $${V_R} \simeq 90\,V$$
27.
An arc lamp requires a direct current of $$10 A$$ at $$80 V$$ to function. If it is connected to a $$220V\left( {rms} \right),50Hz\,AC$$ supply, the series inductor needed for it to work is close to :
A
$$0.044 H$$
B
$$0.065 H$$
C
$$80 H$$
D
$$0.08 H$$
Answer :
$$0.065 H$$
Here
$$i = \frac{e}{{\sqrt {{R^2} + X_L^2} }} = \frac{e}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }} = \frac{e}{{\sqrt {{R^2} + 4{\pi ^2}{v^2}{L^2}} }}$$
$$10 = \frac{{220}}{{\sqrt {64 + 4{\pi ^2}{{\left( {50} \right)}^2}L} }}\,\,\left[ {\because R = \frac{V}{I} = \frac{{80}}{{10}} = 8} \right]$$
On solving we get
$$L = 0.065\,{\text{H}}$$
28.
In a circuit inductance $$L$$ and capacitance $$C$$ are connected as shown in figure. $${A_1}$$ and $${A_2}$$ are ammeters. When key $$K$$ is pressed to complete the circuit, then just after closing key $$\left( K \right),$$ the reading of current will be
A
Zero in both $${A_1}$$ and $${A_2}$$
B
maximum in both $${A_1}$$ and $${A_2}$$
C
zero in $${A_1}$$ and maximum in $${A_2}$$
D
maximum in $${A_1}$$ and zero in $${A_2}$$
Answer :
maximum in $${A_1}$$ and zero in $${A_2}$$
There is no $$DC$$ current in inductive circuit because it does not allow $$DC$$ current to flow and maximum $$DC$$ current in capacitive circuit. Hence, the current is zero in $${A_2}$$ and maximum in $${A_1}.$$
29.
In a $$LCR$$ circuit capacitance is changed from $$C$$ to $$2C.$$ For the resonant frequency to remain unchanged, the inductance should be change from $$L$$ to
30.
In an $$LCR$$ circuit as shown below both switches are open initially. Now switch $${S_1}$$ is closed, $${S_2}$$ kept open. ($$q$$ is charge on the capacitor and $$\tau = RC$$ is Capacitive time constant). Which of the following statements is correct ?
Charge on he capacitor at any time $$t$$ is given by $$q = CV\left( {1 - {e^{\frac{t}{\tau }}}} \right)\,\,{\text{at}}\,\,t = 2\tau $$
$$q = CV\left( {1 - {e^{ - 2}}} \right)$$