Question

When 100 volt dc is applied across a solenoid, a current of 1.0 amp flows in it. When 100 volt ac is applied across the same coil, the current drops to 0.5 amp. If the frequency of ac source is 50 Hz the impedance and inductance of the solenoid are

• A. 200 ohms and 0.5 henry
• B. 100 ohms and 0.86 henry
• C. 200 ohms and 1.0 henry
• D. 100 ohms and 0.93 henry

Answer 1

Answer: (A)

200 ohms and 0.5 henry

Answer 2

When dc is applied $i = \frac{V}{R}$ ⇒ $1 = \frac{100}{R}$ ⇒ R = 100Ω. When ac is applied $i = \frac{V}{Z}$ ⇒ $0.5 = \frac{100}{Z}$ ⇒ Z = 200Ω. Hence$Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{R^{2} + 4\pi^{2}\nu^{2}L^{2}}$ ⇒

$(200)^{2} = (100)^{2} + 4\pi^{2}(50)^{2}L^{2}$ ⇒ L = 0.55H.