An inductor 20 × 10^–3 henry, a capacitor $100 μF$ and a resistor $50 Ω$ are connected in series across a source of emf V = 10 sin 314t. Find the energy dissipated in the circuit in 20 minutes. If resistance is removed from the circuit and the value of inductance is doubled, then find the variation of current with time in the new circuit.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

951 J, 0.52 cos 314 t

951 J, 1.52 cos 315 t

851 J, 2.52 cos 412 t

751 J, 2.55 cos 215 t

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} E = (V_{rms} i_{rms} \cos ϕ) t \\ \Rightarrow E = \frac{V^{2}}{2 z} (\frac{R}{z}) t \Rightarrow E = \frac{V^{2} R}{2 z^{2}} t \\ z = \sqrt{R^{2} + {(ωL - \frac{1}{ωc})}^{2}} = 56.16 Ω \end{array} \Rightarrow E = \frac{10^{2} \times 50}{2 \times 56 . 16^{2}} \times 20 \times 60 = 951 J$

When R removed :

$\cos ϕ = \frac{R}{z} = 0 \Rightarrow ϕ = \frac{π}{2}$

$\begin{array}{l} z = X_{C} - X_{L} = \frac{1}{ω C} - ω (2 L) = 19.28 o h m \\ i = \frac{10}{Z} \sin (ω t + ϕ) \\ \Rightarrow i = \frac{10}{19 .28} \cos ω t (ϕ = 90) \\ i = 0 .518 \cos 314 t \end{array}$