An ac source operating at 59 hz with a maximum voltage of 170 v is connected in series with a resistor (r = 1.2 kω) and an inductor (l = 2.4 h). (a) what is the maximum value of the current in the circuit?

I = V/Z

V = voltage, I = current, Z = impedance

First let's find the total impedance of the circuit.

The impedance of the resistor is:

= R

R = resistance

Given values:

R = 1200Ω

Plug in:

= 1200Ω

The impedance of the inductor is:

= j2πfL

f = source frequency, L = inductance

Given values:

f = 59Hz, L = 2.4H

Plug in:

= j2π(59)(2.4) = j889.7Ω

Add up the individual impedances to get the Z, and convert Z to polar form:

Z = +

Z = 1200 + j889.7

Z = 1494∠36.55°Ω

I = V/Z

Given values:

V = 170∠0°V (assume 0 initial phase)

Z = 1494∠36.55°Ω

I = 170∠0°/1494∠36.55°Ω

I = 0.1138∠-36.55°A

Round the magnitude of I to 2 significant figures and now you have your maximum current:

I = 0.11A

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

connect the batteries in series and hope