A resistance of 5 Ω is connected in series with a pure inductance of 0.01 H to a 100 V, 50 Hz supply. Calculate - a) Impedance b) Current c) Power absorbed

 Given data:

 Resistance, R = 5 Ω,

 Inductance, L = 0.01 H,

 Supply voltage = 100 V

Frequency, f = 50 Hz

To find: Impedance, Z = ?     Current drawn, I = ?     Power absorbed, P = ? 

Solution:

𝐖𝐞 𝐡𝐚𝐯𝐞 𝐈𝐧𝐝𝐮𝐜𝐭𝐢𝐯𝐞 𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐜𝐞, 𝐗_𝐋  = 𝟐𝛑𝐟𝐋 = 𝟐𝛑 × 𝟓𝟎 × 𝟎. 𝟎𝟏 = 𝟑. 𝟏𝟒 Ω

𝐈𝐦𝐩𝐞𝐝𝐚𝐧𝐜𝐞, 𝐙 = √𝐑^𝟐 + 𝐗_𝐋^𝟐 = √𝟓 ^𝟐 + 𝟑. 𝟏𝟒^𝟐 = 𝟓. 𝟗 Ω

𝐂𝐮𝐫𝐫𝐞𝐧𝐭,𝐈 = 𝐕/ 𝐙 = 𝟏𝟎𝟎/ 𝟓. 𝟗 = 𝟏𝟔. 𝟗𝟓 𝐀

𝐏𝐨𝐰𝐞𝐫 𝐚𝐛𝐬𝐨𝐫𝐛𝐞𝐝, 𝐏 = 𝐈^𝟐𝐑 = (𝟏𝟔. 𝟗𝟓)^𝟐 × 𝟓 = 𝟏𝟒𝟑𝟔 W