155236
Two identical incandescent light bulbs are connected as shown in the Figure. When the circuit is connected with an AC voltage source of frequency $f$, which of the following observations will be correct?
1 both bulbs will glow alternatively
2 both bulbs will glow with same brightness provided frequency $\mathrm{f}=\frac{1}{2 \pi \sqrt{(1 / \mathrm{LC})}}$
3 bulb $b_{1}$ will light up initially and goes off, bulb $b_{2}$ will be ON constantly
4 bulb $b_{1}$ will blink and bulb $b_{2}$ will be ON constantly
Explanation:
B The circuit shown above is a parallel resonant circuit. Then frequency is $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ at resonance. We know that, at resonance $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Therefore, equal current will flow through both the bulb $b_{1}$ and $b_{2}$. So, both bulb will glow with same brightness.
VITEEE-2008
Alternating Current
155237
From figure shown below a series LCR circuit connected to a variable frequency $200 \mathrm{~V}$ source. $L=5 H, C=80 \mu F$ and $R=40 \Omega$. Then the source frequency which drive the circuit at resonance is
1 $25 \mathrm{~Hz}$
2 $\frac{25}{\pi} \mathrm{Hz}$
3 $50 \mathrm{~Hz}$
4 $\frac{50}{\pi} \mathrm{Hz}$
Explanation:
B Given that, $\mathrm{L}=5 \mathrm{H}, \mathrm{C}=80 \mu \mathrm{F}=80 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=40 \Omega, \mathrm{V}=200 \text { Volt }$ We know that, frequency at resonance, $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{5 \times 80 \times 10^{-6}}}=\frac{100}{4 \pi}$ $\mathrm{f}=\frac{25}{\pi} \mathrm{Hz}$
VITEEE-2007
Alternating Current
155239
Current in the LCR circuit becomes extremely large when
1 frequency of AC supply is increased
2 frequency of $\mathrm{AC}$ supply is decreased
3 inductive reactance becomes equal to capacitive reactance
4 inductance becomes equal to capacitance
Explanation:
C We know that, Impedance $(\mathrm{Z})=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}}$ According to the question, current in L-C-R circuit becomes maximum. So, for maximum current, impedance should be minimum. $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}}$ For minimum impedance $\left(X_{C}-X_{L}\right)^{2}=0$ $X_{C}=X_{L}$ Hence, inductive reactance becomes equal to capacitive reactance.
VITEEE-2006
Alternating Current
155240
In a series L-C-R circuit, an alternating emf (V) and current (i) are given by the equation $\mathrm{V}$ $=V_{0} \sin \omega t, I=I_{0} \sin \left(\omega t+\frac{\pi}{3}\right)$ The average power dissipated in the circuit over a cycle of $\mathrm{AC}$ is :
155236
Two identical incandescent light bulbs are connected as shown in the Figure. When the circuit is connected with an AC voltage source of frequency $f$, which of the following observations will be correct?
1 both bulbs will glow alternatively
2 both bulbs will glow with same brightness provided frequency $\mathrm{f}=\frac{1}{2 \pi \sqrt{(1 / \mathrm{LC})}}$
3 bulb $b_{1}$ will light up initially and goes off, bulb $b_{2}$ will be ON constantly
4 bulb $b_{1}$ will blink and bulb $b_{2}$ will be ON constantly
Explanation:
B The circuit shown above is a parallel resonant circuit. Then frequency is $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ at resonance. We know that, at resonance $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Therefore, equal current will flow through both the bulb $b_{1}$ and $b_{2}$. So, both bulb will glow with same brightness.
VITEEE-2008
Alternating Current
155237
From figure shown below a series LCR circuit connected to a variable frequency $200 \mathrm{~V}$ source. $L=5 H, C=80 \mu F$ and $R=40 \Omega$. Then the source frequency which drive the circuit at resonance is
1 $25 \mathrm{~Hz}$
2 $\frac{25}{\pi} \mathrm{Hz}$
3 $50 \mathrm{~Hz}$
4 $\frac{50}{\pi} \mathrm{Hz}$
Explanation:
B Given that, $\mathrm{L}=5 \mathrm{H}, \mathrm{C}=80 \mu \mathrm{F}=80 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=40 \Omega, \mathrm{V}=200 \text { Volt }$ We know that, frequency at resonance, $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{5 \times 80 \times 10^{-6}}}=\frac{100}{4 \pi}$ $\mathrm{f}=\frac{25}{\pi} \mathrm{Hz}$
VITEEE-2007
Alternating Current
155239
Current in the LCR circuit becomes extremely large when
1 frequency of AC supply is increased
2 frequency of $\mathrm{AC}$ supply is decreased
3 inductive reactance becomes equal to capacitive reactance
4 inductance becomes equal to capacitance
Explanation:
C We know that, Impedance $(\mathrm{Z})=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}}$ According to the question, current in L-C-R circuit becomes maximum. So, for maximum current, impedance should be minimum. $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}}$ For minimum impedance $\left(X_{C}-X_{L}\right)^{2}=0$ $X_{C}=X_{L}$ Hence, inductive reactance becomes equal to capacitive reactance.
VITEEE-2006
Alternating Current
155240
In a series L-C-R circuit, an alternating emf (V) and current (i) are given by the equation $\mathrm{V}$ $=V_{0} \sin \omega t, I=I_{0} \sin \left(\omega t+\frac{\pi}{3}\right)$ The average power dissipated in the circuit over a cycle of $\mathrm{AC}$ is :
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155236
Two identical incandescent light bulbs are connected as shown in the Figure. When the circuit is connected with an AC voltage source of frequency $f$, which of the following observations will be correct?
1 both bulbs will glow alternatively
2 both bulbs will glow with same brightness provided frequency $\mathrm{f}=\frac{1}{2 \pi \sqrt{(1 / \mathrm{LC})}}$
3 bulb $b_{1}$ will light up initially and goes off, bulb $b_{2}$ will be ON constantly
4 bulb $b_{1}$ will blink and bulb $b_{2}$ will be ON constantly
Explanation:
B The circuit shown above is a parallel resonant circuit. Then frequency is $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ at resonance. We know that, at resonance $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Therefore, equal current will flow through both the bulb $b_{1}$ and $b_{2}$. So, both bulb will glow with same brightness.
VITEEE-2008
Alternating Current
155237
From figure shown below a series LCR circuit connected to a variable frequency $200 \mathrm{~V}$ source. $L=5 H, C=80 \mu F$ and $R=40 \Omega$. Then the source frequency which drive the circuit at resonance is
1 $25 \mathrm{~Hz}$
2 $\frac{25}{\pi} \mathrm{Hz}$
3 $50 \mathrm{~Hz}$
4 $\frac{50}{\pi} \mathrm{Hz}$
Explanation:
B Given that, $\mathrm{L}=5 \mathrm{H}, \mathrm{C}=80 \mu \mathrm{F}=80 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=40 \Omega, \mathrm{V}=200 \text { Volt }$ We know that, frequency at resonance, $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{5 \times 80 \times 10^{-6}}}=\frac{100}{4 \pi}$ $\mathrm{f}=\frac{25}{\pi} \mathrm{Hz}$
VITEEE-2007
Alternating Current
155239
Current in the LCR circuit becomes extremely large when
1 frequency of AC supply is increased
2 frequency of $\mathrm{AC}$ supply is decreased
3 inductive reactance becomes equal to capacitive reactance
4 inductance becomes equal to capacitance
Explanation:
C We know that, Impedance $(\mathrm{Z})=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}}$ According to the question, current in L-C-R circuit becomes maximum. So, for maximum current, impedance should be minimum. $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}}$ For minimum impedance $\left(X_{C}-X_{L}\right)^{2}=0$ $X_{C}=X_{L}$ Hence, inductive reactance becomes equal to capacitive reactance.
VITEEE-2006
Alternating Current
155240
In a series L-C-R circuit, an alternating emf (V) and current (i) are given by the equation $\mathrm{V}$ $=V_{0} \sin \omega t, I=I_{0} \sin \left(\omega t+\frac{\pi}{3}\right)$ The average power dissipated in the circuit over a cycle of $\mathrm{AC}$ is :
155236
Two identical incandescent light bulbs are connected as shown in the Figure. When the circuit is connected with an AC voltage source of frequency $f$, which of the following observations will be correct?
1 both bulbs will glow alternatively
2 both bulbs will glow with same brightness provided frequency $\mathrm{f}=\frac{1}{2 \pi \sqrt{(1 / \mathrm{LC})}}$
3 bulb $b_{1}$ will light up initially and goes off, bulb $b_{2}$ will be ON constantly
4 bulb $b_{1}$ will blink and bulb $b_{2}$ will be ON constantly
Explanation:
B The circuit shown above is a parallel resonant circuit. Then frequency is $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ at resonance. We know that, at resonance $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Therefore, equal current will flow through both the bulb $b_{1}$ and $b_{2}$. So, both bulb will glow with same brightness.
VITEEE-2008
Alternating Current
155237
From figure shown below a series LCR circuit connected to a variable frequency $200 \mathrm{~V}$ source. $L=5 H, C=80 \mu F$ and $R=40 \Omega$. Then the source frequency which drive the circuit at resonance is
1 $25 \mathrm{~Hz}$
2 $\frac{25}{\pi} \mathrm{Hz}$
3 $50 \mathrm{~Hz}$
4 $\frac{50}{\pi} \mathrm{Hz}$
Explanation:
B Given that, $\mathrm{L}=5 \mathrm{H}, \mathrm{C}=80 \mu \mathrm{F}=80 \times 10^{-6} \mathrm{~F}$ $\mathrm{R}=40 \Omega, \mathrm{V}=200 \text { Volt }$ We know that, frequency at resonance, $\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ $\mathrm{f}=\frac{1}{2 \pi \sqrt{5 \times 80 \times 10^{-6}}}=\frac{100}{4 \pi}$ $\mathrm{f}=\frac{25}{\pi} \mathrm{Hz}$
VITEEE-2007
Alternating Current
155239
Current in the LCR circuit becomes extremely large when
1 frequency of AC supply is increased
2 frequency of $\mathrm{AC}$ supply is decreased
3 inductive reactance becomes equal to capacitive reactance
4 inductance becomes equal to capacitance
Explanation:
C We know that, Impedance $(\mathrm{Z})=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}}$ According to the question, current in L-C-R circuit becomes maximum. So, for maximum current, impedance should be minimum. $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^{2}}$ For minimum impedance $\left(X_{C}-X_{L}\right)^{2}=0$ $X_{C}=X_{L}$ Hence, inductive reactance becomes equal to capacitive reactance.
VITEEE-2006
Alternating Current
155240
In a series L-C-R circuit, an alternating emf (V) and current (i) are given by the equation $\mathrm{V}$ $=V_{0} \sin \omega t, I=I_{0} \sin \left(\omega t+\frac{\pi}{3}\right)$ The average power dissipated in the circuit over a cycle of $\mathrm{AC}$ is :
