Verified Solution • 2025–26 Session
moderate
QNEET-2018-Ph-894: NEET 2018 Physics - Class 12 NEET-2018-Ph-894 Solution: An inductor 20 mH a capacitor \(100 F\) and a resistor \(50 \)…
An inductor 20 mH a capacitor and a resistor are connected in series across a source of emf V = 10 sin 314 t. The power loss in the circuit is 0.79 W 0.43 W 2.74 W 1.13 W
Multiple Choice Options
- A 0.79 W
- B 0.43 W
- C 2.74 W
- D 1.13 W
Step-by-Step Verified Solution
Average power in impedance
Z = R 2 + ( X C _ X L ) 2
where X C
= capacitive
reactance and X L
= inductive reactance.
Also X C = 1 _ C and X L = _ L
_ Z = ( 50 ) 2 + ( 1 314 _ 100 _ 10 _ 6 _ 314 _ 20 _ 10 _ 3 ) 2
_ Z = 56.15 �
I rms = V r m s Z = V m 2 _ Z = 10 2 _ 56
Hence power loss in the circuit
= ( 10 2 _ 56 ) 2 _ 50 = 0.79 W