A resistance of 10 ohm is joined in series with an inductance of 0.5 henry. What capacitance should be put in series with the combination to obtain the maximum current? What will be the potential difference across the resistance, inductance and capacitor? The current is being supplied by 200 volts and 50 cycles per second mains.

Solution

The current in the circuit would be maximum when X_L = Xc

i.e.,_{$ωL = \frac{1}{ωC}, or C = \frac{1}{ω^{2} L}$}Therefore_,

C = \frac{1}{(2 πf)^{2} L} = \frac{1}{(2 \times 3.14 \times 50)^{2} \times 0.5} = 20.24 \times 10^{- 6} farad .

Here,

ωL = \frac{1}{ωC} .

So the impedance Z of the circuit,

Z = \sqrt{[R^{2} + {(ωL - \frac{1}{ωC})}^{2}]} and R = 10 ohm

I = \frac{E}{R} = \frac{200}{10} = 20 amp .

Potential difference across resistance

V_R = I × R
= 20 × 10
= 200 volt

Potential difference across inductance

V_{L} = ωL \times I = (2 π \times 50 \times 0.5) \times 20 = 3142 volt .

Potential difference across condenser

V_{C} = \frac{1}{ωC} = I \times ωL = 3142 volt .

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