In a series R L C circuit, potential differences across R, L and C are 30 V , 60 V and 100 V respectively as shown in figure. The e.m.f. of source (in volts) is

Question

In a series R L C circuit, potential differences across R, L and C are 30 V , 60 V and 100 V respectively as shown in figure. The e.m.f. of source (in volts) is (A) 190 (B) 70 (C) 50 (D) 40

Tardigrade Admin · Accepted Answer

emf =V textrms =√VR2+(VL-Vc)2 =√(30)2+(100-60)2 =√(30)2+(40)2=50 V

Q. In a series $R L C$ circuit, potential differences across $R, L$ and $C$ are $30 V, 60 V$ and $100 V$ respectively as shown in figure. The e.m.f. of source (in volts) is

190

70

50

40

emf $= V_{rms} = V_{R}^{2} + (V_{L} - V_{c})^{2}$
$= (30)^{2} + (100 - 60)^{2}$
$= (30)^{2} + (40)^{2} = 50 V$

Q. In a series RLC circuit, potential differences across R,L and C are 30V,60V and 100V respectively as shown in figure. The e.m.f. of source (in volts) is

190

70

50

40

emf =Vrms ​=VR2​+(VL​−Vc​)2​ =(30)2+(100−60)2​ =(30)2+(40)2​=50V

Q. In a series $R L C$ circuit, potential differences across $R, L$ and $C$ are $30 V, 60 V$ and $100 V$ respectively as shown in figure. The e.m.f. of source (in volts) is

emf $= V_{rms} = V_{R}^{2} + (V_{L} - V_{c})^{2}$
$= (30)^{2} + (100 - 60)^{2}$
$= (30)^{2} + (40)^{2} = 50 V$