{"id":2763,"date":"2024-08-25T11:32:49","date_gmt":"2024-08-25T11:32:49","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2763"},"modified":"2024-08-25T11:32:49","modified_gmt":"2024-08-25T11:32:49","slug":"phasor-or-vector-method","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/25\/phasor-or-vector-method\/","title":{"rendered":"\u00a0\u00a0PHASOR (OR VECTOR) METHOD"},"content":{"rendered":"\n

To solve parallel AC circuits by this method, we proceed as follows:<\/p>\n\n\n\n

Step I: <\/strong>Draw the circuit as per the given problem, as shown in Figure 7.47(a)<\/a> (here, for illustration, we have considered two branches connected in parallel. One branch contains resistance and inductance in series, whereas second branch contains resistance and capacitance in series. The supply voltage is V<\/em> V).<\/p>\n\n\n\n

Step II: <\/strong>Find the impedance of each branch of the circuit separately.<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n
\"img\"\/<\/figure>\n\n\n\n

Fig. 7.47<\/strong>  (a) Circuit diagram (b) Phasor diagram<\/p>\n\n\n\n

<\/a>Step III: <\/strong>Determine the magnitude of current and phase angle with the voltage in each branch.<\/p>\n\n\n\n

\"img\"; \u0278<\/em>1<\/sub>\"img\"; (lagging) (for inductive branch)<\/p>\n\n\n\n

\"img\"; \u0278<\/em>2<\/sub>\"img\"; (lagging) (for capacitive branch)<\/p>\n\n\n\n

Step IV: <\/strong>Draw the phasor diagram by considering voltage as the reference phasor. Represent the branch currents on it as shown in Figure 7.47(b)<\/a>.<\/p>\n\n\n\n

Step V: <\/strong>Find the phasor sum of branch currents by the method of components.<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>XX<\/sub> = I<\/em>1<\/sub> cos \u0278<\/em>1<\/sub> + I<\/em>2<\/sub> cos \u0278<\/em>2<\/sub><\/p>\n\n\n\n

I<\/em>YY<\/sub> = \u2212 I<\/em>1<\/sub> sin \u0278<\/em>1<\/sub> + I<\/em>2<\/sub> sin \u0278<\/em>2<\/sub> (negative)<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Step VI: <\/strong>Find the phase angle \u0278<\/em> between the total current I<\/em> and circuit voltage V<\/em>.<\/p>\n\n\n\n

\"img\" lagging (since I<\/em>YY <\/sub>is negative)<\/p>\n\n\n\n

Power factor of the circuit=cos\u0278<\/em>(lagging)<\/p>\n\n\n\n

or<\/p>\n\n\n\n

\"img\" (lagging)<\/p>\n\n\n\n

Example 7.30<\/strong><\/p>\n\n\n\n

A coil of resistance 15 \u03a9 and inductance 0.05 H is connected in parallel with a non\u2212inductive resistance of 20 \u03a9. Find (i) current in each branch of the circuit, (ii) total current supplied, (iii) phase angle and pf of combination when a voltage of 200 V at 50 Hz is applied, and (iv) power consumed in the circuit.<\/p>\n\n\n\n

Solution:<\/em><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Fig. 7.48<\/strong>  (a) Circuit as per data (b) Phasor diagram<\/p>\n\n\n\n

Applied voltage, V<\/em> = 200 V; supply frequency,f<\/em> = 50 Hz<\/p>\n\n\n\n

<\/a>Branch I<\/em><\/strong><\/p>\n\n\n\n

 <\/p>\n\n\n\n

R<\/em>2<\/sub> = 15 \u03a9; X<\/em>L1<\/sub> = 2 \u03c0<\/em> fL<\/em>1<\/sub> = 2\u03c0<\/em> \u00d7 50 \u00d7 0.05 = 15.7 \u03a9<\/p>\n\n\n\n

Impedance,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Current in the coil, \"image\"<\/p>\n\n\n\n

Phase angle,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Branch II<\/em><\/strong><\/p>\n\n\n\n

Resistance,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

R<\/em>2<\/sub> = 20 \u03a9<\/p>\n\n\n\n

Branch current,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Phase angle,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u02782<\/sub><\/em>=0(I<\/em>2<\/sub> is in phase with V<\/em>)<\/p>\n\n\n\n

The two currents are shown vectorially in Figure 7.48(b)<\/a>. Resolving the currents horizontally and vertically,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>XX<\/sub> = I<\/em>2<\/sub> + I<\/em>1<\/sub> cos \u0278<\/em>1<\/sub> = 10 + 9.2 cos 46.3\u00b0 = 10 + 9.2 \u00d7 6,909 = 16.356 A<\/p>\n\n\n\n

I<\/em>yy<\/sub>= 0 \u2212 I<\/em>1<\/sub>sin\u0278<\/em>1<\/sub>= 0 \u2212 9.2sin 46.3\u00b0 = \u22129.2\u00d70.723 = \u22126.65A<\/p>\n\n\n\n

Total current supplied,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Phase angle,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Power factor of the circuit,cos \u0278 = cos(\u221222.126\u00b0) = 0.9264(lagging)<\/p>\n\n\n\n

Power,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

P = V<\/em>I<\/em> cos \u0278<\/em> = 200 \u00d7 17.659 \u00d7 0.9264 = 3,271.3W<\/p>\n\n\n\n

Example 7.31<\/strong><\/p>\n\n\n\n

A series AC circuit has a resistance of 15 \u03a9 and inductive reactance of 10 \u03a9. Calculate the value of capacitor that is connected across this series combination so that system has unity power factor. The frequency of AC supply is 50 Hz.<\/p>\n\n\n\n

(U.P.T.U. 2005\u201306)<\/strong><\/p>\n\n\n\n

Solution:<\/em><\/p>\n\n\n\n

The circuit is shown in Figure 7.49(a)<\/a>. Let the supply voltage be V<\/em> V.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Fig. 7.49<\/strong>  (a) Circuit as per data (b) Phasor diagram<\/p>\n\n\n\n

<\/a>Branch I<\/em><\/strong><\/p>\n\n\n\n

Impedance, \"image\"<\/p>\n\n\n\n

Current, \"image\"<\/p>\n\n\n\n

Power factor, \"image\"<\/p>\n\n\n\n

Phase angle, \u0278<\/em> = cos\u22121<\/sup>0.832 = 33.7\u00b0 lag<\/p>\n\n\n\n

Reactive component of current,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Branch II<\/em><\/strong><\/p>\n\n\n\n

Current drawn by the capacitor, \"image\"<\/p>\n\n\n\n

Power factor of the circuit will be unity when<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>c<\/sub> = I<\/em>r1<\/sub><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Now,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Example 7.32<\/strong><\/p>\n\n\n\n

The parallel circuit shown in Figure 7.50(a)<\/a> is connected across a single phase 100 V, 50 Hz AC supply. Calculate the (i) branch currents, (ii) total current, (iii) supply power factor, and (iv) active and reactive power supplied by the supply.<\/p>\n\n\n\n

(U.P.T.U. 2006\u221207)<\/strong><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Fig. 7.50<\/strong>  (a) Given circuit diagram (b) Circuit as per data<\/p>\n\n\n\n

<\/a>Solution:<\/em><\/p>\n\n\n\n

 <\/p>\n\n\n\n

V <\/em>= 100 V, f<\/em> = 50 Hz<\/p>\n\n\n\n

The circuit is shown in Figure 7.48(a)<\/a>.<\/p>\n\n\n\n