A circuit takes a current of 3 A at a power factor of 0.6 lagging when connected to a 150 V, 50 Hz supply. Another circuit takes a current of 5 A at a power factor of 0.8 leading when connected to the same supply. If these two circuits are connected in series across 230 V, 50 Hz supply. Calculate the current and power consumed

 

Given data:

 For Circuit-1 Current, I= 3 A,

 Power factor, Cos Φ = 0.6 lagging

Supply voltage, V=150 V,

Supply frequency, f = 50 Hz

 For Circuit-2 Currrent, I= 5 A,

 Power factor, Cos Φ = 0.8 leading

 Supply voltage, V=150 V,

Supply frequency, f = 50 Hz

 To find:

 Current, I = ?

Power, P = ?

Solution:

 𝐈𝐦𝐩𝐞𝐝𝐚𝐧𝐜𝐞, 𝐙_𝟏 = 𝐕/ 𝐈_𝟏 = 𝟏𝟓𝟎/ 𝟑 = 𝟓𝟎 Ω

𝐑𝐞𝐬𝐢𝐬𝐭𝐚𝐧𝐜𝐞, 𝐑_𝟏 = 𝐙_𝟏 𝐂𝐨𝐬Ф = 𝟓𝟎 × 𝟎. 𝟔 = 𝟑𝟎 Ω

𝐈𝐧𝐝𝐮𝐜𝐭𝐢𝐯𝐞 𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐜𝐞, 𝐗_𝐋 = √𝐙^𝟐 − 𝐑^𝟐 = √𝟓𝟎^𝟐 − 𝟑𝟎^𝟐 = 𝟒𝟎 Ω

 𝐑𝟏 = 𝟑𝟎 Ω 𝐚𝐧𝐝 𝐗_𝐋 = 𝟒𝟎 Ω

 𝐈𝐦𝐩𝐞𝐝𝐚𝐧𝐜𝐞, 𝐙_𝟐 = 𝐕/ 𝐈_𝟐 = 𝟏𝟓𝟎/ 𝟓 = 𝟑𝟎 Ω

 𝐑𝐞𝐬𝐢𝐬𝐭𝐚𝐧𝐜𝐞, 𝐑_𝟐 = 𝐙_𝟐 𝐂𝐨𝐬Ф = 𝟑𝟎 × 𝟎. 𝟖 = 𝟐𝟒 Ω

 𝐂𝐚𝐩𝐚𝐜𝐢𝐭𝐢𝐯𝐞 𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐜𝐞, 𝐗_𝐂 = √𝐙 ^𝟐 − 𝐑^𝟐 = √𝟑𝟎^𝟐 − 𝟐𝟒^𝟐 = 𝟏𝟖 Ω

 𝐑_𝟐 = 𝟐𝟒 Ω 𝐚𝐧𝐝 𝐗_𝐂 = 𝟏𝟖 Ω

When both of them are connected in series, we have

 𝐑_𝐄𝐐 = 𝐑_𝟏 + 𝐑_𝟐 = 𝟑𝟎 + 𝟐𝟒 = 𝟓𝟒 Ω

𝐗_𝐄𝐐 = 𝐗_𝐋 − 𝐗_𝐂 = 𝟒𝟎 − 𝟏𝟖 = 𝟐𝟐 Ω

 𝐄𝐪𝐮𝐢𝐯𝐚𝐥𝐞𝐧𝐭 𝐈𝐦𝐩𝐞𝐝𝐚𝐧𝐜𝐞, 𝐙_𝐄𝐐 = √𝐑_𝐄𝐐^𝟐 + 𝐗_𝐄𝐐^𝟐 = √𝟓𝟒^𝟐 + 𝟐𝟐^𝟐 = 𝟓𝟖. 𝟑𝟏 Ω

 𝐂𝐮𝐫𝐫𝐞𝐧𝐭,𝐈 = 𝐕/ 𝐙 = 𝟐𝟑𝟎/ 𝟓𝟖. 𝟑𝟏 = 𝟑. 𝟗𝟒 𝐀

 𝐏𝐨𝐰𝐞𝐫 𝐜𝐨𝐧𝐬𝐮𝐦𝐞𝐝, 𝐏 = 𝐈^𝟐𝐑 = (𝟑. 𝟗𝟒)^𝟐 × 𝟓𝟒 = 𝟖𝟒𝟎W