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Question
A resistor of 50 Ω, a capacitor of `(25/pi)` µF and an inductor of `(4/pi)` H are connected in series across an ac source whose voltage (in volts) is given by V = 70 sin (100 πt). Calculate:
- the net reactance of the circuit
- the impedance of the circuit
- the effective value of current in the circuit.
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Solution
Given:
Resistance (R) = 50 Ω
Capacitor (C) = `25/pi`
Inductance (L) = `4/pi`
V = 70 sin (100 πt)
From this equation, we can identify ω as 100 πt and V as 70 volts.
(a) We can find net reactance from the following formula.
Net reactance, XL + XC
XL = ωL (ω = 100 π, L = 4/π)
`X_L = 100pi xx 4/pi`
`X_L = 400 Omega`
`X_C = 1/(omegaC)` (ω = 100 π, C = 25/π)
`X_C = 1/((100pi xx 25/pi xx 10^-6))`
`X_C = 10^4/25`
`X_C = 400Omega`
Net reactance = (400 + 400) Ω = 800 Ω
(b) Impedance of the circuit
We can find the impedance of the circuit from the following formula:
`Z = sqrt(R^2 + (X_L - X_C)^2)`
Where R is the resistance.
Z = `sqrt((50)^2 + [(400 - 400)^2])`
Z = 50 Ω
(c) Effective value of current
I = `V/Z` (V0 = 70 volt, Z = 50 Ω)
I = `70/(50sqrt2)`
I ≅ 1 A
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