Question

When a resistor is connected across the terminals of an ac generator (113 V) that has a fixed frequency, there is a current of 0.324 A in the resistor. When an inductor is connected across the terminals of the same generator, there is a current of 0.461 A in the inductor. When both the resistor and the inductor are connected in series between the terminals of this generator, what is (a) the impedance of the series combination and (b) the phase angle between the current and the voltage of the generator

Answer 1

a) Z = 426.3 Ω

b) Ф = 35.1°

Step-by-step explanation:

The ac generator has a voltage(Vrms) of 113 V, when connected to a resistor there is a current(Irms) of 0.324 A. The resistance of the resistor(R) is given by:

`R=(V_(rms) )/(I_(rms) )`

Substituting values:

`R=(113 )/(0.324 )=348.765`

R = 348.765 Ω

When the generator is connector to an inductor the current(Irms) is 0.461 A.

The impedance of the inductor XL is given by:

`X_(l)=(V_(rms) )/(I_(rms) )`

Substituting values:

`X_(l)=(113 )/(0.461 )=245.12`

XL = 245.12 Ω

When both the resistor and the inductor are connected in series between the terminals of this generator

(a) the impedance of the series combination

the impedance of the series combination(Z) =
`\sqrt{R^(2)+X_(L )^(2) }`

`Z=\sqrt{R^(2)+X_(L )^(2) }=\sqrt{348.765^(2)+245.12^(2) }` = 426.3 Ω

Z = 426.3 Ω

(b) the phase angle between the current and the voltage of the generator

the phase angle(Ф) is given by:

tanФ = XL/R

Ф = tan⁻¹(XL/R)

Ф = tan⁻¹(245.12/348.765) = tan⁻¹(0.7028) = 35.1°

Ф = 35.1°