Use phasor techniques to determine the current supplied by the source given that V = 6 <0o v, R = 7 Ω, C = 12 μF, L = 4 mH and ω = 2000 rad/sec. The current supplied by the source is I = ∠ ° A

Given Information:

Voltage = V = 6 < 0° V

Resistance = R = 7 Ω

Capacitance = C = 12 μF

Inductance = L = 4 mH

Frequency = ω = 2000 rad/sec

Required Information:

Current = I = ?

I = 0.174 <78.25° A

Explanation:

The current can be found using

I = V/Z

Where Z is the impedance an can be calculated as

Z = R + J(XL - XC)

Where XL is inductive reactance and XC is capacitive reactance

XL = ω*L

XL = 2000*0.004

XL = 8 Ω

XC = 1/ω*C

XC = 1/2000*12x10⁻⁶

XC = 41.66 Ω

So the impedance can be calculated now

Z = R + J(XL - XC)

Z = 7 + J(8 - 41.66)

Z = 7 - J33.66 Ω

|Z| = √(7²) + (-33.66)²

|Z| = 34.38 Ω

Magnitude of the current

I = V/|Z|

I = 6/34.38

I = 0.174 A

Magnitude and angle both

I = 6 < 0°/(7 - J33.66)

I = 0.0355 + j0.1708 A

in the polar form

I = 0.174 <78.25° A

the warning light flashes for a second or two when the engine starts. if it stays on, there's must be something wrong with the coolant. you should never dare to continue driving when the light is on. pull over as soon as possible and shut the engine off.

hope this is what you're talking about