You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

RLC Circuit

When an induc...

Question

When an inductor, a capacitor and a resistor are connected in series across a source of alterating emf, the potential difference across the inductor, capacitor and resistor are 70V, 30V and 30V respectively. Find the emf of the source

50V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

60V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

80V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 50V
$V = \sqrt{V_{R}^{2} + [V_{L} - V_{c}]^{2}}$
$= \sqrt{30^{2} + [70 - 30]^{2}}$
$= \sqrt{30^{2} + 40^{2}} = \sqrt{50^{2}}$
=50 V

Suggest Corrections

Similar questions

Q. A series

L C R

circuit is connected to a source of alternating emf

50 V

and if the potential differences across inductor and capacitor are

90 V

and

60 V

respectively, the potential difference across resistor is:

Q. An inductor

20 m H

, a capacitor

50 μ F

and a resistor

40 Ω

are connected in series across a source of emf

V = 10 sin 340 t

. The power loss in A.C. circuit is:

Q. A resistor of

40 Ω

, an inductor of

5 / π H

and a capacitor of

50 / π μ F

are connected in series across a source of alternating voltage of

140 sin 100 π t

Volts. Find the r.m.s voltage across the resistor, the inductor and the capacitor. Is algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox

Q. An inductor

20

mH, a capacitor

100

μ

F and a resistor

50

Ω

are connected in series across a source of emf, V

= 10

sin 314

t. The power loss in the circuit is?

Q. An inductor of reactance

60 Ω

, capacitor of reactance

30 Ω

and resistor of resistance

40 Ω

are connected in series across a source of emf

V = 100 sin 628 t

. Find power loss in the circuit.

Join BYJU'S Learning Program

When an inductor, a capacitor and a resistor are connected in series across a source of alterating emf, the potential difference across the inductor, capacitor and resistor are 70V, 30V and 30V respectively. Find the emf of the source

The correct option is A 50VV=√V2R+[VL−Vc]2 =√302+[70−30]2 =√302+402=√502 =50 V

The correct option is A 50V
$V = \sqrt{V_{R}^{2} + [V_{L} - V_{c}]^{2}}$
$= \sqrt{30^{2} + [70 - 30]^{2}}$
$= \sqrt{30^{2} + 40^{2}} = \sqrt{50^{2}}$
=50 V