Question

When 4 V DC is connected across an inductor current is 0.2 A. When AC of 4 V is applied the current is 0.1 A. Then self inductance of the coil is ……….; w = 1000 rads^–1 –

• A. 20 mH
• B. 40 mH
• C. 20$\sqrt { 3 }$mH
• D. 20$\sqrt { 3 }$mH

Answer 1

Answer: (C)

20$\sqrt { 3 }$mH

Answer 2

Z = $\frac { 4 } { 0.1 }$ =40W = $\sqrt { \mathrm { R } ^ { 2 } + ( \omega \mathrm { L } ) ^ { 2 } }$ \R=$\frac { 4 } { 0.2 }$= 20 W

Lw = 20$\sqrt { 3 }$ Mh