Home
Class 12
PHYSICS
A circuit element shown in the figureure...

A circuit element shown in the figureure as a box is having either a capacitor or an inductor. The power factor of the circuit is `0.8`, while current lags behind the voltage. Find

(a) the source voltage V,
(b) the nature of the element in box and find its value.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
(a) `V_R=IR=80V, V_C=100V`
and `V_L=IX_X=160V`
`:. V=sqrt(V_R^2+(V_L-V_C)^2)=100V`
(b) since the current is lagging behind, there should be an inductor in the box.
`X_C=V_C/I=100Omega`
Now `0.8=R/Z80/sqrt((80)^2+(X_L-100)^2)`
solving we get
`X_L=160Omega`
or `omega L=160`
`:. (2pifL)=160`
`:. L=160/(2pif)=160/((2pi)xx50)`
`=1.6/piH`.
Promotional Banner

Topper's Solved these Questions

  • ALTERNATING CURRENT

    DC PANDEY ENGLISH|Exercise JEE MAIN|62 Videos

  • ALTERNATING CURRENT

    DC PANDEY ENGLISH|Exercise Level 2 Comprehension|6 Videos

  • ATOMS

    DC PANDEY ENGLISH|Exercise MEDICAL ENTRANCES GALLERY|42 Videos

Similar Questions

Explore conceptually related problems

A circuit shown in the figure contains a box having either a capacitor or an inductor. The power factor of the circuit is 0.8, while current lags behind the voltage. Then find the inductance/capacitance of the box is S.I. unit (take pi=3.2 ). The circuit draws 1 ampere current.

In the circuit shown in figure power factor of box is 0.5 and power factor of circuit is sqrt3/2 .Current leading the voltage. Find the effective resistance of the box.

In the given circuit, power factor between A and B is 0.5 and over all power factor of the circuit is 1. Find the value of Rand L

An inductive circuit draws a power 550 W from a 220V-50Hz source. The power factor of the circuit is 0.8. The current in the circuit lags behind the voltage. Show that a capacitor of about (1)/(42 pi) xx10^(-2) F will have to be commected in the circuit to bring its power factor to unity.

A circuit draws a power of 550 watt from a source of 220 volt, 50 Hz . The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. To make the power factor of the circuit as 1.0, The capacitance should be connected in series with it is

In a series LCR circuit R= 200(Omega) and the voltage and the frequency of the main supply is 220V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30(@) . On taking out the inductor from the circuit the current leads the voltage by 30(@) . The power dissipated in the LCR circuit is

A circuit draws 330 W from a 110 V , 60 Hz AC line. The power factor is 0.6 and the current lags the voltage. The capacitance of a series capacitor that will result in a power factor of unity is equal to

An LCR circuit contains resistance of 100 Omega and a supply of 200V at 300 rads^(-1) angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by 60^(@) . If, on the other hand , only inductor is taken out,the current leads the voltage by 60^(o) . The current flowing in the circuit is

In a series C-R circuit shown in figureure, the applied voltage is 10 V and the voltage across capacitor is found to 8 V . The voltage across R , and the phase difference between current and the applied voltage will respectively be .

In a series C-R circuit shown in figureure, the applied voltage is 10 V and the voltage across capacitor is found to 8 V . The voltage across R , and the phase difference between current and the applied voltage will respectively be .