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प्रश्न
An alternating emf of 220 V is applied to a circuit containing a resistor R having the resistance of 160Ω and a capacitor ‘C’ is series. The current is found to lead the supply voltage by an angle
`θ = tan^(-1) ("3/4")`
a) Calculate
1) The capacitive reactance
2) The impedance of the circuit
3) Current flowing in the circuit
b) If the frequency of the applied emf is 50 Hz, what is the value of the capacitance of the capacitor ‘C’?
उत्तर
a) E = 220 V, R = 160, C = Capacitor
Current leads voltage by `phi = tan^(-1) (3/4)`
Find `X_C, Z, I`
a) Given θ = `tan^(-1) (3/4)` ...(1)
but θ = `tan^(-1) ((X_C)/R)` ...(2)
From 1 and 2
tan θ = 3/4
but tan θ = `X_C/R`
`:. 3/4 = X_C/R`
∵ R = 160
`:. 3/4= X_C/160`
`:. X_C = 3/4 xx 160`
`X_C = 120 Ω`
Impedance Z = `sqrt(R^2 + X_(C^2))`
`= [(160)^2 + (120)^2]^("1/2")`
Z = 200 Ω
`I_0 = V/Z = 220/2000 = 1.1A => I_0 = 1.1 A`
b) f = 50 Hz
`X_C = 120Ω`
`X_C = 1/(2pifC)`
`:. C = 1/((2pif)xx(X_C)) = 1/(2pi xx 50xx 120) = 2.65 xx10^(-5) F = 26.5 muF`
`C = 26.5 mu F`
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