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Question
A series RL circuit with R = 10 Ω and L = `(100/pi)` mH is connected to an ac source of voltage V = 141 sin (100 πt), where V is in volts and t is in seconds. Calculate
- the impedance of the circuit
- phase angle, and
- the voltage drop across the inductor.
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Solution
Given:
R = 10 Ω
L = `(100/pi)` mH
V = 141 sin (100 π)t
From this equation, we get the value of ω = 100π and V = 141 volt
- To find: Impedance (Z)
Z = `sqrt(R^2 + X_L^2)`
Where Z is the impedance, R is the resistance, and XL is the impedance,
XL = ωL
XL = `(100pi xx 100)/(pi xx 10^-3)`
XL = 10Ω
Z = `sqrt(R^2 + X_L^2)`
Z = `sqrt((10)^2 + (10)^2)`
Z = `sqrt200`
Z = `10sqrt2`Ω - Phase Angle (Φ):
We can calculate the phase angle by the following formula,
`cosphi = R/Z`
`cosphi = 10/(10sqrt2)`
`cosphi = 1/sqrt2`
`phi = 45^circ` - Voltage drop:
`V_L = IX_L`
= `V/Z xx X_L`
`V_L = 141/(10sqrt2) xx 10`
`V_L = 141/sqrt2`
`V_L ≅ 100` volt
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