First slide

Electrical Resonance series circuit

Question

In a series circuit, $C = 2 μF, L = 1 mH and R = 10 Ω .$ When the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be

Moderate

A

1 : 1
B

1 : 2
C

2 : 1
D

1 : 5

Solution

Current will be maximum in the condition of resonance, so $i_{max} = \frac{E}{R} = \frac{E}{10} A$
Energy stored in the coil, $W_{L} = \frac{1}{2} {Li}_{max}^{2} = \frac{1}{2} L {(\frac{E}{10})}^{2}$
$= \frac{1}{2} \times 10^{- 3} (\frac{E^{2}}{100}) = \frac{1}{2} \times 10^{- 5} E^{2} joule$
Energy stored in the capacitor,
$W_{C} = \frac{1}{2} {CE}^{2} = \frac{1}{2} \times 2 \times 10^{- 6} E^{2} = 10^{- 6} E^{2} joule$
$∴ \frac{W_{C}}{W_{L}} = \frac{1}{5}$

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