Home
Class 12
PHYSICS
A resistor of 200 Omegaand a capacitor o...

A resistor of `200 Omega`and a capacitor of `15.0muF` are connected in series to a `220V`,` 50Hz` source.
(a) Calculate the current in the circuit .
(b) Calcutalte the voltage (rms) across the resistor and the inductor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.

Text Solution

Verified by Experts

Here, `R = 200 ohm`,
`C = 15.0 mu F = 15 xx 10^(-6) F`
`E_(v) = 220 V, v = 50 Hz, I_(v) = ?`
`V_(R ) = ?, V_(C ) = ?`
Now `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)`
`= (1)/(2 xx 3.14 xx 50 xx 15 xx 10^(-6)) = 212.3 Omega`
(a) Impedance of the circuit,
`Z = sqrt(R^(2) + X_(C)^(2))`
`= sqrt(2000^(2) + (212.3)^(2))`
`= 291.7 ohm`
`:.` Current in the circuit,
`I_(v) = (E_(v))/(Z) = (220)/(291.7) = 0.75 A`
(b) `V_(R ) = I_(v) xx R = 0.75 xx 200 = 150.8 V`
`V_(C ) - I_(v) X_(C ) = 0.75 xx 212.3 = 159.2 V`
`V_(R ) + V_(C ) = 150.8 + 159.2 = 310 V`, which is more than the source voltage of 220 V. This paradox is resolved by the fact that the two voltages are not in same phase. Therefore, they connot be added like ordinary number. As `V_(R )` and `V_(C )` are out of phase by `90^(@)`, therefore,
`V_(RC) = sqrt(V_(R )^(2) + V_(C )^(2)) = sqrt((150.8)^(2) + (159.2)^(2))`
`= 220 V`, the source voltage.
Promotional Banner

Topper's Solved these Questions

  • ELECTROMAGNETIC INDUCTION & ALTERNATING CURRENT

    PRADEEP|Exercise Solved Examples (b)|1 Videos

  • ELECTROMAGNETIC INDUCTION & ALTERNATING CURRENT

    PRADEEP|Exercise Short Answer Qusetions|2 Videos

  • DUAL NATURE OF RADIATION AND MATTER

    PRADEEP|Exercise Exercise|191 Videos

  • ELECTROMAGNETIC WAVES

    PRADEEP|Exercise II Focus multiple choice question|5 Videos

Similar Questions

Explore conceptually related problems

A resistor of 200 Omega and a capacitor of 15.0(mu)F are connected in series to a 220V, 50Hz source. (a) Calculate the current in the circuit . (b) Calcutalte the voltage (rms) across the resistor and the inductor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.

An a.c. source generating a voltage v=v+(m) sin omega t is connected to a capacitor of capacitor of capacitance C. Find the expression for the current, I, flowing through it. Plot a graph of v and I versus omega t to show that the current is pi//2 ahead of the voltage. A resistor of 200 Omega and a capacitor of 15.0 mu F are connected in series to a 220 V, 50 Hz a.c. source. Calculate the current in the circuit and the rms voltage across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltages ? If yes, resolve the paradox.

A 31.4Omega resistor and 0.1H inductor are connected in series to a 200V, 50Hz ac source. Calculate (i) the current in the circuit

In a series RC circuit, R = 30 ohm, C = 0.25 mu F , V = 100 V, omega = 10000 rad//s . Find the current in the circuit and calculate the voltage across the resistor and capacitor. Is the algebraic sum of these voltages more than the source voltage ? If yes, resolve the paradox.

A resistance of 200Omega and capacitor of 15muF are connected in series to a 220 V, 50 Hz AC source. The voltage (rms) across the resistor and capacitor is that

A 0.2 k Omega resistor and 15 mu F capacitor are connected in series to a 220 V, 50 Hz ac source. The impadance of the circuit is

A resistor of resistance 400 ohm and a capacitor of reactance 200 ohm are connected in series ot a 200 V, 50 Hz source. If the current in the circuit is 0.49 A, find the voltage across the resistor and capacitor. What is the value of inductance required so that ane current are in phase ?

A 0.21 H inductor and a 12 ohm resistance are connected in series to a 220 V . 50 Hz ac source. Calculate the current in the circuit and the phase angle between the current and the source voltage.

A 31.4Omega resistor and 0.1H inductor are connected in series to a 200V, 50Hz ac source. Calculate (ii) the voltage (rms) across the inductor and the resistor.