A 1.07 H inductor is connected in series with a fluorescent lamp to limit the current drawn by the lamp. If the combination is connected to a 28.9 Hz, 170 V line, and if the voltage across the lamp is to be 17.6 V, what is the current in the circuit

0.784 A

Explanation:

From the question,

Note that the current in the circuit is the same as the current flowing through the inductor since they are both connected in series.

I = VL/XL Equation 1

Where I = current flowing through the circuit, VL = Voltage drop across the inductor,  XL = reactance of the inductor.

XL = 2πfL Equation 2

Given: f = 28.9 Hz, L = 1.07 H, π = 3.143

XL = 2(3.143)(28.9)(1.07)

XL = 194.38 Ω.

VL = V-Vf

VL = 170-17.6

VL = 152.4 V

Substitute these values into equation 1

I = 152.4/194.38

I = 0.784 A

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures