Answer:
i. 43.5 mH ii. 16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°
Explanation:
i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω
The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.
So, x = 2π(50)L = 100πL Ω = 314.16L Ω
Since the current is the same when the 240 V supply is applied, then
the impedance Z = √(R² + X²) = 240 V/15 A
√(R² + X²) = 16 Ω
8.33² + X² = 16²
69.3889 + X² = 256
X² = 256 - 69.3889
X² = 186.6111
X = √186.6111
X = 13.66 Ω
Since X = 314.16L = 13.66 Ω
L = 13.66/314.16
= 0.0435 H
= 43.5 mH
ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.
So in phasor form Z = (8.33 + j13.66) Ω
iii. The phase difference θ between the current and voltage is
θ = tan⁻¹X/R
= tan⁻¹(314.16L/R)
= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)
= tan⁻¹(13.66/8.33)
= tan⁻¹(1.6406)
= 58.64°
rotational kinetic energy of it's particles
The resistance in this circuit is 39.8 ohms.
Explanation:
Any circuit having resistor, battery and ammeter connected in series will obey the ohm's law in basic case. So according to the Ohm's law, the current flowing in the circuit through the ammeter will be equal to the voltage shown in the voltmeter or battery and resistor is the proportionality constant. So with this law
V = IR
So, Resistance R = V/I
As the voltage is given as 23.90 V and the current is given as 0.6 A, then resistance is
R = 23.90/0.6 = 39.8 ohms.
So, the resistance in this circuit is 39.8 ohms.
Because its just enough to where its not out f the gravitational pull and not close enough to be pulled back to earth. Hope it helps<span />
