QBManager

PHXII07:ALTERNATING CURRENT

355988 A resistor and a capacitor are connected in series with an $a$ . $c$ . source. If the potential drop across the capacitor is 8 $V$ and that across resistor is 6 $V$ , the applied voltage is

1

17 V

2

10 V

3

7 V

4

9 V

Explanation:

Let the applied voltage be $V$ volt.
Here, $V_{R} = 6 V, V_{C} = 8 V$
$V = \sqrt{V_{R}^{2} + V_{C}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = 10 V$

PHXII07:ALTERNATING CURRENT

355989 In a series $R - C$ circuit shown in figure, the applied voltage is $10 V$ and the voltage across capacitor is found to be $8 V$ . Then, the voltage across $R$ and the phase difference between current and the applied voltage will respectively be
supporting img

1

6 V, \tan^{- 1} (\frac{4}{3})

2

3 V, \tan^{- 1} (\frac{3}{4})

3

6 V, \tan^{- 1} (\frac{5}{3})

4 None of these

Explanation:

We know that, for series $R - C$ circuit
$V^{2} = V_{C}^{2} + V_{R}^{2}$
supporting img
$100 = 64 + V_{R}^{2}$
$\Rightarrow V_{R}^{2} = 36 \Rightarrow V_{R} = \sqrt{36} = 6 V$
Also, $\tan ϕ = \frac{V_{C}}{V_{R}} \Rightarrow \tan ϕ = \frac{8}{6}$
$\tan ϕ = \frac{4}{3}$
$∴ ϕ = \tan^{- 1} (\frac{4}{3})$

AIIMS - 2018

PHXII07:ALTERNATING CURRENT

355990 The diagram shows a capacitor $C$ and a resistor $R$ connected in series to an ac source. $V_{1}$ and $V_{2}$ are voltmeters and $A$ is an ammeter
Consider the fallowing statements
I.
Reading in $A$ and $V_{2}$ are always in phase.
II.
Reading $V_{1}$ is ahead in phase with reading in $V_{2}$ .
III.
Reading in $A$ and $V_{1}$ are always in phase.
Which of these statements are/is correct.
supporting img

1 II and III only

2 I and II only

3 I only

4 II only

Explanation:

Phasor diagram of an $R C$ circuit is
supporting img

So only I is correct

PHXII07:ALTERNATING CURRENT

355991 In $R - C$ circuit, $ω = 100 r a d s^{- 1}, R = 100 Ω, C = 20 μ F$ .What is impedance?

1

510 Ω

2

200 Ω

3

250 Ω

4

300 Ω

Explanation:

Impedance for an $R$ - $C$ circuit is
given by
$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}$
$= \sqrt{{(100)}^{2} + {(\frac{1}{100 \times 20 \times 10^{- 6}})}^{2}}$
$= \sqrt{10^{4} + 25 \times 10^{4}}$
$= 100 \sqrt{26} = 510 Ω$

PHXII07:ALTERNATING CURRENT

355992 A $100 Ω$ resistance and a capacitor of $100 Ω$ reactance are connected in series across a 220 $V$ source. When the capacitor is 50% charged, the peak value of the displacement current is

1

2.2 A

2

11 A

3

4.4 A

4

11 \sqrt{2} A

Explanation:

$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{100^{2} + 100^{2}} = 100 \sqrt{2}$
$i_{0} = \frac{V_{0}}{Z} = \frac{V_{r m s} \sqrt{2}}{Z} = \frac{220 \sqrt{2}}{100 \sqrt{2}} = 2.2 A$

NEET - 2016

PHXII07:ALTERNATING CURRENT

355988 A resistor and a capacitor are connected in series with an $a$ . $c$ . source. If the potential drop across the capacitor is 8 $V$ and that across resistor is 6 $V$ , the applied voltage is

1

17 V

2

10 V

3

7 V

4

9 V

Explanation:

Let the applied voltage be $V$ volt.
Here, $V_{R} = 6 V, V_{C} = 8 V$
$V = \sqrt{V_{R}^{2} + V_{C}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = 10 V$

PHXII07:ALTERNATING CURRENT

355989 In a series $R - C$ circuit shown in figure, the applied voltage is $10 V$ and the voltage across capacitor is found to be $8 V$ . Then, the voltage across $R$ and the phase difference between current and the applied voltage will respectively be
supporting img

1

6 V, \tan^{- 1} (\frac{4}{3})

2

3 V, \tan^{- 1} (\frac{3}{4})

3

6 V, \tan^{- 1} (\frac{5}{3})

4 None of these

Explanation:

We know that, for series $R - C$ circuit
$V^{2} = V_{C}^{2} + V_{R}^{2}$
supporting img
$100 = 64 + V_{R}^{2}$
$\Rightarrow V_{R}^{2} = 36 \Rightarrow V_{R} = \sqrt{36} = 6 V$
Also, $\tan ϕ = \frac{V_{C}}{V_{R}} \Rightarrow \tan ϕ = \frac{8}{6}$
$\tan ϕ = \frac{4}{3}$
$∴ ϕ = \tan^{- 1} (\frac{4}{3})$

AIIMS - 2018

PHXII07:ALTERNATING CURRENT

355990 The diagram shows a capacitor $C$ and a resistor $R$ connected in series to an ac source. $V_{1}$ and $V_{2}$ are voltmeters and $A$ is an ammeter
Consider the fallowing statements
I.
Reading in $A$ and $V_{2}$ are always in phase.
II.
Reading $V_{1}$ is ahead in phase with reading in $V_{2}$ .
III.
Reading in $A$ and $V_{1}$ are always in phase.
Which of these statements are/is correct.
supporting img

1 II and III only

2 I and II only

3 I only

4 II only

Explanation:

Phasor diagram of an $R C$ circuit is
supporting img

So only I is correct

PHXII07:ALTERNATING CURRENT

355991 In $R - C$ circuit, $ω = 100 r a d s^{- 1}, R = 100 Ω, C = 20 μ F$ .What is impedance?

1

510 Ω

2

200 Ω

3

250 Ω

4

300 Ω

Explanation:

Impedance for an $R$ - $C$ circuit is
given by
$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}$
$= \sqrt{{(100)}^{2} + {(\frac{1}{100 \times 20 \times 10^{- 6}})}^{2}}$
$= \sqrt{10^{4} + 25 \times 10^{4}}$
$= 100 \sqrt{26} = 510 Ω$

PHXII07:ALTERNATING CURRENT

355992 A $100 Ω$ resistance and a capacitor of $100 Ω$ reactance are connected in series across a 220 $V$ source. When the capacitor is 50% charged, the peak value of the displacement current is

1

2.2 A

2

11 A

3

4.4 A

4

11 \sqrt{2} A

Explanation:

$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{100^{2} + 100^{2}} = 100 \sqrt{2}$
$i_{0} = \frac{V_{0}}{Z} = \frac{V_{r m s} \sqrt{2}}{Z} = \frac{220 \sqrt{2}}{100 \sqrt{2}} = 2.2 A$

NEET - 2016

PHXII07:ALTERNATING CURRENT

355988 A resistor and a capacitor are connected in series with an $a$ . $c$ . source. If the potential drop across the capacitor is 8 $V$ and that across resistor is 6 $V$ , the applied voltage is

1

17 V

2

10 V

3

7 V

4

9 V

Explanation:

Let the applied voltage be $V$ volt.
Here, $V_{R} = 6 V, V_{C} = 8 V$
$V = \sqrt{V_{R}^{2} + V_{C}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = 10 V$

PHXII07:ALTERNATING CURRENT

355989 In a series $R - C$ circuit shown in figure, the applied voltage is $10 V$ and the voltage across capacitor is found to be $8 V$ . Then, the voltage across $R$ and the phase difference between current and the applied voltage will respectively be
supporting img

1

6 V, \tan^{- 1} (\frac{4}{3})

2

3 V, \tan^{- 1} (\frac{3}{4})

3

6 V, \tan^{- 1} (\frac{5}{3})

4 None of these

Explanation:

We know that, for series $R - C$ circuit
$V^{2} = V_{C}^{2} + V_{R}^{2}$
supporting img
$100 = 64 + V_{R}^{2}$
$\Rightarrow V_{R}^{2} = 36 \Rightarrow V_{R} = \sqrt{36} = 6 V$
Also, $\tan ϕ = \frac{V_{C}}{V_{R}} \Rightarrow \tan ϕ = \frac{8}{6}$
$\tan ϕ = \frac{4}{3}$
$∴ ϕ = \tan^{- 1} (\frac{4}{3})$

AIIMS - 2018

PHXII07:ALTERNATING CURRENT

355990 The diagram shows a capacitor $C$ and a resistor $R$ connected in series to an ac source. $V_{1}$ and $V_{2}$ are voltmeters and $A$ is an ammeter
Consider the fallowing statements
I.
Reading in $A$ and $V_{2}$ are always in phase.
II.
Reading $V_{1}$ is ahead in phase with reading in $V_{2}$ .
III.
Reading in $A$ and $V_{1}$ are always in phase.
Which of these statements are/is correct.
supporting img

1 II and III only

2 I and II only

3 I only

4 II only

Explanation:

Phasor diagram of an $R C$ circuit is
supporting img

So only I is correct

PHXII07:ALTERNATING CURRENT

355991 In $R - C$ circuit, $ω = 100 r a d s^{- 1}, R = 100 Ω, C = 20 μ F$ .What is impedance?

1

510 Ω

2

200 Ω

3

250 Ω

4

300 Ω

Explanation:

Impedance for an $R$ - $C$ circuit is
given by
$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}$
$= \sqrt{{(100)}^{2} + {(\frac{1}{100 \times 20 \times 10^{- 6}})}^{2}}$
$= \sqrt{10^{4} + 25 \times 10^{4}}$
$= 100 \sqrt{26} = 510 Ω$

PHXII07:ALTERNATING CURRENT

355992 A $100 Ω$ resistance and a capacitor of $100 Ω$ reactance are connected in series across a 220 $V$ source. When the capacitor is 50% charged, the peak value of the displacement current is

1

2.2 A

2

11 A

3

4.4 A

4

11 \sqrt{2} A

Explanation:

$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{100^{2} + 100^{2}} = 100 \sqrt{2}$
$i_{0} = \frac{V_{0}}{Z} = \frac{V_{r m s} \sqrt{2}}{Z} = \frac{220 \sqrt{2}}{100 \sqrt{2}} = 2.2 A$

NEET - 2016

PHXII07:ALTERNATING CURRENT

355988 A resistor and a capacitor are connected in series with an $a$ . $c$ . source. If the potential drop across the capacitor is 8 $V$ and that across resistor is 6 $V$ , the applied voltage is

1

17 V

2

10 V

3

7 V

4

9 V

Explanation:

Let the applied voltage be $V$ volt.
Here, $V_{R} = 6 V, V_{C} = 8 V$
$V = \sqrt{V_{R}^{2} + V_{C}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = 10 V$

PHXII07:ALTERNATING CURRENT

355989 In a series $R - C$ circuit shown in figure, the applied voltage is $10 V$ and the voltage across capacitor is found to be $8 V$ . Then, the voltage across $R$ and the phase difference between current and the applied voltage will respectively be
supporting img

1

6 V, \tan^{- 1} (\frac{4}{3})

2

3 V, \tan^{- 1} (\frac{3}{4})

3

6 V, \tan^{- 1} (\frac{5}{3})

4 None of these

Explanation:

We know that, for series $R - C$ circuit
$V^{2} = V_{C}^{2} + V_{R}^{2}$
supporting img
$100 = 64 + V_{R}^{2}$
$\Rightarrow V_{R}^{2} = 36 \Rightarrow V_{R} = \sqrt{36} = 6 V$
Also, $\tan ϕ = \frac{V_{C}}{V_{R}} \Rightarrow \tan ϕ = \frac{8}{6}$
$\tan ϕ = \frac{4}{3}$
$∴ ϕ = \tan^{- 1} (\frac{4}{3})$

AIIMS - 2018

PHXII07:ALTERNATING CURRENT

355990 The diagram shows a capacitor $C$ and a resistor $R$ connected in series to an ac source. $V_{1}$ and $V_{2}$ are voltmeters and $A$ is an ammeter
Consider the fallowing statements
I.
Reading in $A$ and $V_{2}$ are always in phase.
II.
Reading $V_{1}$ is ahead in phase with reading in $V_{2}$ .
III.
Reading in $A$ and $V_{1}$ are always in phase.
Which of these statements are/is correct.
supporting img

1 II and III only

2 I and II only

3 I only

4 II only

Explanation:

Phasor diagram of an $R C$ circuit is
supporting img

So only I is correct

PHXII07:ALTERNATING CURRENT

355991 In $R - C$ circuit, $ω = 100 r a d s^{- 1}, R = 100 Ω, C = 20 μ F$ .What is impedance?

1

510 Ω

2

200 Ω

3

250 Ω

4

300 Ω

Explanation:

Impedance for an $R$ - $C$ circuit is
given by
$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}$
$= \sqrt{{(100)}^{2} + {(\frac{1}{100 \times 20 \times 10^{- 6}})}^{2}}$
$= \sqrt{10^{4} + 25 \times 10^{4}}$
$= 100 \sqrt{26} = 510 Ω$

PHXII07:ALTERNATING CURRENT

355992 A $100 Ω$ resistance and a capacitor of $100 Ω$ reactance are connected in series across a 220 $V$ source. When the capacitor is 50% charged, the peak value of the displacement current is

1

2.2 A

2

11 A

3

4.4 A

4

11 \sqrt{2} A

Explanation:

$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{100^{2} + 100^{2}} = 100 \sqrt{2}$
$i_{0} = \frac{V_{0}}{Z} = \frac{V_{r m s} \sqrt{2}}{Z} = \frac{220 \sqrt{2}}{100 \sqrt{2}} = 2.2 A$

NEET - 2016

PHXII07:ALTERNATING CURRENT

355988 A resistor and a capacitor are connected in series with an $a$ . $c$ . source. If the potential drop across the capacitor is 8 $V$ and that across resistor is 6 $V$ , the applied voltage is

1

17 V

2

10 V

3

7 V

4

9 V

Explanation:

Let the applied voltage be $V$ volt.
Here, $V_{R} = 6 V, V_{C} = 8 V$
$V = \sqrt{V_{R}^{2} + V_{C}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = 10 V$

PHXII07:ALTERNATING CURRENT

355989 In a series $R - C$ circuit shown in figure, the applied voltage is $10 V$ and the voltage across capacitor is found to be $8 V$ . Then, the voltage across $R$ and the phase difference between current and the applied voltage will respectively be
supporting img

1

6 V, \tan^{- 1} (\frac{4}{3})

2

3 V, \tan^{- 1} (\frac{3}{4})

3

6 V, \tan^{- 1} (\frac{5}{3})

4 None of these

Explanation:

We know that, for series $R - C$ circuit
$V^{2} = V_{C}^{2} + V_{R}^{2}$
supporting img
$100 = 64 + V_{R}^{2}$
$\Rightarrow V_{R}^{2} = 36 \Rightarrow V_{R} = \sqrt{36} = 6 V$
Also, $\tan ϕ = \frac{V_{C}}{V_{R}} \Rightarrow \tan ϕ = \frac{8}{6}$
$\tan ϕ = \frac{4}{3}$
$∴ ϕ = \tan^{- 1} (\frac{4}{3})$

AIIMS - 2018

PHXII07:ALTERNATING CURRENT

355990 The diagram shows a capacitor $C$ and a resistor $R$ connected in series to an ac source. $V_{1}$ and $V_{2}$ are voltmeters and $A$ is an ammeter
Consider the fallowing statements
I.
Reading in $A$ and $V_{2}$ are always in phase.
II.
Reading $V_{1}$ is ahead in phase with reading in $V_{2}$ .
III.
Reading in $A$ and $V_{1}$ are always in phase.
Which of these statements are/is correct.
supporting img

1 II and III only

2 I and II only

3 I only

4 II only

Explanation:

Phasor diagram of an $R C$ circuit is
supporting img

So only I is correct

PHXII07:ALTERNATING CURRENT

355991 In $R - C$ circuit, $ω = 100 r a d s^{- 1}, R = 100 Ω, C = 20 μ F$ .What is impedance?

1

510 Ω

2

200 Ω

3

250 Ω

4

300 Ω

Explanation:

Impedance for an $R$ - $C$ circuit is
given by
$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}$
$= \sqrt{{(100)}^{2} + {(\frac{1}{100 \times 20 \times 10^{- 6}})}^{2}}$
$= \sqrt{10^{4} + 25 \times 10^{4}}$
$= 100 \sqrt{26} = 510 Ω$

PHXII07:ALTERNATING CURRENT

355992 A $100 Ω$ resistance and a capacitor of $100 Ω$ reactance are connected in series across a 220 $V$ source. When the capacitor is 50% charged, the peak value of the displacement current is

1

2.2 A

2

11 A

3

4.4 A

4

11 \sqrt{2} A

Explanation:

$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{100^{2} + 100^{2}} = 100 \sqrt{2}$
$i_{0} = \frac{V_{0}}{Z} = \frac{V_{r m s} \sqrt{2}}{Z} = \frac{220 \sqrt{2}}{100 \sqrt{2}} = 2.2 A$

NEET - 2016

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: