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Alternating Current

155259 IN an L-C-R circuit, if impedance is $\sqrt{3}$ times of resistance and capacitive reactance is zero. Find the phase difference.

1 Zero

2

30^{\circ}

3

60^{\circ}

4 data is incomplete

Explanation:

C
We know that,
$\tan ϕ = \frac{X_{L} - X_{C}}{R}$
$X_{C} = 0, X_{L} = \sqrt{3} R$
$\tan ϕ = \frac{\sqrt{3} R}{R}$
$\tan ϕ = \sqrt{3}$
$ϕ = \tan^{- 1} (\sqrt{3})$
$ϕ = 60^{\circ}$

UP CPMT-2003

Alternating Current

155260 A resistor $R$ , and inductor $L$ and a capacitor $C$ are connected in series to a source of frequency $n$ . If the resonant frequency is $n_{r}$ , then the current lags behind voltage when :

1

n = 0

2 n

< n_{r}

3

n = n_{r}

4

n > n_{r}

Explanation:

D When the reactance of inductance is more than reactance of condenser then current lag behind the voltage -
$X_{L} > X_{C}$
$ω L < \frac{1}{ω C}$
$Or ω > \frac{1}{\sqrt{LC}}$
Then, $n > \frac{1}{2 π \sqrt{LC}}$
Or $n > n_{r}$ ( $n_{r}$ is resonant frequency)

SRMJEEE - 2007

Alternating Current

155261 For a series LCR circuit, the rms values of voltage across various components are $V_{L} = 90 V, V_{C} = 60 V$ and $V_{R} = 40 V$ Volt, The rms value of the voltage applied to the circuit is :

1

190 V

2

110 V

3

70 V

4

50 V

Explanation:

D Given that,
Voltage across inductor $(V_{L}) = 90 V$
Voltage across capacitor $(V_{C}) = 60 V$
Voltage across resistance $(V_{R}) = 40 V$
$Now, V_{rms} = \sqrt{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}$
$V_{rms} = \sqrt{(40)^{2} + (90 - 60)^{2}}$
$V_{rms} = \sqrt{(40)^{2} + (30)^{2}}$
$V_{rms} = 50 V$
Hence, the rms value of the voltage applied to the circuit is $50 V$ .

MP PMT-2013

Alternating Current

155262 In a series $L - C - R$ circuit, resistance $R = 10 Ω$ and the impedance $Z = 10 Ω$ . The phase difference between the current and the voltage is

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

A Given that,
Resistance $(R) = 10 Ω$
Impedance $(Z) = 10 Ω$
We know that, phase difference -
$\cos ϕ = \frac{R}{Z} = \frac{10}{10}$
$\cos ϕ = 1$
$ϕ = \cos^{- 1} (1)$
$ϕ = 0^{\circ}$

JIPMER-2010

Alternating Current

155259 IN an L-C-R circuit, if impedance is $\sqrt{3}$ times of resistance and capacitive reactance is zero. Find the phase difference.

1 Zero

2

30^{\circ}

3

60^{\circ}

4 data is incomplete

Explanation:

C
We know that,
$\tan ϕ = \frac{X_{L} - X_{C}}{R}$
$X_{C} = 0, X_{L} = \sqrt{3} R$
$\tan ϕ = \frac{\sqrt{3} R}{R}$
$\tan ϕ = \sqrt{3}$
$ϕ = \tan^{- 1} (\sqrt{3})$
$ϕ = 60^{\circ}$

UP CPMT-2003

Alternating Current

155260 A resistor $R$ , and inductor $L$ and a capacitor $C$ are connected in series to a source of frequency $n$ . If the resonant frequency is $n_{r}$ , then the current lags behind voltage when :

1

n = 0

2 n

< n_{r}

3

n = n_{r}

4

n > n_{r}

Explanation:

D When the reactance of inductance is more than reactance of condenser then current lag behind the voltage -
$X_{L} > X_{C}$
$ω L < \frac{1}{ω C}$
$Or ω > \frac{1}{\sqrt{LC}}$
Then, $n > \frac{1}{2 π \sqrt{LC}}$
Or $n > n_{r}$ ( $n_{r}$ is resonant frequency)

SRMJEEE - 2007

Alternating Current

155261 For a series LCR circuit, the rms values of voltage across various components are $V_{L} = 90 V, V_{C} = 60 V$ and $V_{R} = 40 V$ Volt, The rms value of the voltage applied to the circuit is :

1

190 V

2

110 V

3

70 V

4

50 V

Explanation:

D Given that,
Voltage across inductor $(V_{L}) = 90 V$
Voltage across capacitor $(V_{C}) = 60 V$
Voltage across resistance $(V_{R}) = 40 V$
$Now, V_{rms} = \sqrt{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}$
$V_{rms} = \sqrt{(40)^{2} + (90 - 60)^{2}}$
$V_{rms} = \sqrt{(40)^{2} + (30)^{2}}$
$V_{rms} = 50 V$
Hence, the rms value of the voltage applied to the circuit is $50 V$ .

MP PMT-2013

Alternating Current

155262 In a series $L - C - R$ circuit, resistance $R = 10 Ω$ and the impedance $Z = 10 Ω$ . The phase difference between the current and the voltage is

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

A Given that,
Resistance $(R) = 10 Ω$
Impedance $(Z) = 10 Ω$
We know that, phase difference -
$\cos ϕ = \frac{R}{Z} = \frac{10}{10}$
$\cos ϕ = 1$
$ϕ = \cos^{- 1} (1)$
$ϕ = 0^{\circ}$

JIPMER-2010

Alternating Current

155259 IN an L-C-R circuit, if impedance is $\sqrt{3}$ times of resistance and capacitive reactance is zero. Find the phase difference.

1 Zero

2

30^{\circ}

3

60^{\circ}

4 data is incomplete

Explanation:

C
We know that,
$\tan ϕ = \frac{X_{L} - X_{C}}{R}$
$X_{C} = 0, X_{L} = \sqrt{3} R$
$\tan ϕ = \frac{\sqrt{3} R}{R}$
$\tan ϕ = \sqrt{3}$
$ϕ = \tan^{- 1} (\sqrt{3})$
$ϕ = 60^{\circ}$

UP CPMT-2003

Alternating Current

155260 A resistor $R$ , and inductor $L$ and a capacitor $C$ are connected in series to a source of frequency $n$ . If the resonant frequency is $n_{r}$ , then the current lags behind voltage when :

1

n = 0

2 n

< n_{r}

3

n = n_{r}

4

n > n_{r}

Explanation:

D When the reactance of inductance is more than reactance of condenser then current lag behind the voltage -
$X_{L} > X_{C}$
$ω L < \frac{1}{ω C}$
$Or ω > \frac{1}{\sqrt{LC}}$
Then, $n > \frac{1}{2 π \sqrt{LC}}$
Or $n > n_{r}$ ( $n_{r}$ is resonant frequency)

SRMJEEE - 2007

Alternating Current

155261 For a series LCR circuit, the rms values of voltage across various components are $V_{L} = 90 V, V_{C} = 60 V$ and $V_{R} = 40 V$ Volt, The rms value of the voltage applied to the circuit is :

1

190 V

2

110 V

3

70 V

4

50 V

Explanation:

D Given that,
Voltage across inductor $(V_{L}) = 90 V$
Voltage across capacitor $(V_{C}) = 60 V$
Voltage across resistance $(V_{R}) = 40 V$
$Now, V_{rms} = \sqrt{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}$
$V_{rms} = \sqrt{(40)^{2} + (90 - 60)^{2}}$
$V_{rms} = \sqrt{(40)^{2} + (30)^{2}}$
$V_{rms} = 50 V$
Hence, the rms value of the voltage applied to the circuit is $50 V$ .

MP PMT-2013

Alternating Current

155262 In a series $L - C - R$ circuit, resistance $R = 10 Ω$ and the impedance $Z = 10 Ω$ . The phase difference between the current and the voltage is

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

A Given that,
Resistance $(R) = 10 Ω$
Impedance $(Z) = 10 Ω$
We know that, phase difference -
$\cos ϕ = \frac{R}{Z} = \frac{10}{10}$
$\cos ϕ = 1$
$ϕ = \cos^{- 1} (1)$
$ϕ = 0^{\circ}$

JIPMER-2010

Alternating Current

155259 IN an L-C-R circuit, if impedance is $\sqrt{3}$ times of resistance and capacitive reactance is zero. Find the phase difference.

1 Zero

2

30^{\circ}

3

60^{\circ}

4 data is incomplete

Explanation:

C
We know that,
$\tan ϕ = \frac{X_{L} - X_{C}}{R}$
$X_{C} = 0, X_{L} = \sqrt{3} R$
$\tan ϕ = \frac{\sqrt{3} R}{R}$
$\tan ϕ = \sqrt{3}$
$ϕ = \tan^{- 1} (\sqrt{3})$
$ϕ = 60^{\circ}$

UP CPMT-2003

Alternating Current

155260 A resistor $R$ , and inductor $L$ and a capacitor $C$ are connected in series to a source of frequency $n$ . If the resonant frequency is $n_{r}$ , then the current lags behind voltage when :

1

n = 0

2 n

< n_{r}

3

n = n_{r}

4

n > n_{r}

Explanation:

D When the reactance of inductance is more than reactance of condenser then current lag behind the voltage -
$X_{L} > X_{C}$
$ω L < \frac{1}{ω C}$
$Or ω > \frac{1}{\sqrt{LC}}$
Then, $n > \frac{1}{2 π \sqrt{LC}}$
Or $n > n_{r}$ ( $n_{r}$ is resonant frequency)

SRMJEEE - 2007

Alternating Current

155261 For a series LCR circuit, the rms values of voltage across various components are $V_{L} = 90 V, V_{C} = 60 V$ and $V_{R} = 40 V$ Volt, The rms value of the voltage applied to the circuit is :

1

190 V

2

110 V

3

70 V

4

50 V

Explanation:

D Given that,
Voltage across inductor $(V_{L}) = 90 V$
Voltage across capacitor $(V_{C}) = 60 V$
Voltage across resistance $(V_{R}) = 40 V$
$Now, V_{rms} = \sqrt{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}$
$V_{rms} = \sqrt{(40)^{2} + (90 - 60)^{2}}$
$V_{rms} = \sqrt{(40)^{2} + (30)^{2}}$
$V_{rms} = 50 V$
Hence, the rms value of the voltage applied to the circuit is $50 V$ .

MP PMT-2013

Alternating Current

155262 In a series $L - C - R$ circuit, resistance $R = 10 Ω$ and the impedance $Z = 10 Ω$ . The phase difference between the current and the voltage is

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

A Given that,
Resistance $(R) = 10 Ω$
Impedance $(Z) = 10 Ω$
We know that, phase difference -
$\cos ϕ = \frac{R}{Z} = \frac{10}{10}$
$\cos ϕ = 1$
$ϕ = \cos^{- 1} (1)$
$ϕ = 0^{\circ}$

JIPMER-2010

Question Bank Series

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