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:

1. the net reactance of the circuit
2. the impedance of the circuit
3. the effective value of current in the circuit.

Answer 1

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, X_L + X_C 

X_L = 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