Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 1)
                   
                                       
                                1.
A 1.5 k
 resistor and a coil with a 2.2 k
 inductive reactance are in series across an 18 V ac source. The power factor is
 
                                    
 resistor and a coil with a 2.2 k
 inductive reactance are in series across an 18 V ac source. The power factor isDiscussion:
20 comments Page 1 of 2.
                
                        Siddharth said: 
                         
                        7 years ago
                
                Power factor = R ÷ Z.
So, R = 1.3k ohm = 1.3 * 10^3 =1300 ohm,
inductive reactians FL = 2.2k ohm =2.2 * 10^3,
= 2200 ohm.
&. Z = (?).
Z = √(R^2 + FL^2),
Z = √(1500^2 + 2200^2),
Z= 2663.
& Power factor = R ÷ Z,
= 1500 ÷ 2663,
= 0.5632.
So, ans is (0.564).
                So, R = 1.3k ohm = 1.3 * 10^3 =1300 ohm,
inductive reactians FL = 2.2k ohm =2.2 * 10^3,
= 2200 ohm.
&. Z = (?).
Z = √(R^2 + FL^2),
Z = √(1500^2 + 2200^2),
Z= 2663.
& Power factor = R ÷ Z,
= 1500 ÷ 2663,
= 0.5632.
So, ans is (0.564).
                     (3)
                
            
                        Noel Zamora said: 
                         
                        1 decade ago
                
                tan 2.2/1.5 = 1.4667, arc tan 1.4667 = 55.71 deg., 
Therefore cos 55.71 deg. = 0.564 or pf = 0.564(lagging).
                Therefore cos 55.71 deg. = 0.564 or pf = 0.564(lagging).
                        Hadiro tafoua said: 
                         
                        1 decade ago
                
                Using all method above we get close intervals of answers. Since it all corresponds to electric quantities.
                
                        Pavan said: 
                         
                        2 years ago
                
                @All.
The main simple thing we need to see here is the power factor should not be greater than 1.
                The main simple thing we need to see here is the power factor should not be greater than 1.
                     (3)
                
            
                        Yousuf said: 
                         
                        1 decade ago
                
                |z|=sqrt(1500^2+2200^2)=2663
Angle=inverse tan(2200\1500)=55.7
Power Factor=Cos(55.7)=0.564
                Angle=inverse tan(2200\1500)=55.7
Power Factor=Cos(55.7)=0.564
                        AMIT KUMAR JAISWAL said: 
                         
                        1 decade ago
                
                POWER FACTOR = (R/Z).
Z = SQRT(1.5^2 +2.2^2).
Z = 2.662.
P.F. = 1.5/2.662 = .563.
                Z = SQRT(1.5^2 +2.2^2).
Z = 2.662.
P.F. = 1.5/2.662 = .563.
                        Karimulla said: 
                         
                        1 decade ago
                
                Generally power factor i.e. cos(angle b/w v,i) always less than one.
                
                        VENKATESH said: 
                         
                        1 decade ago
                
                R = 1.5.
Xl = 2.2.
Z = Root of R square+xl square.
Cos = R/Z.
                Xl = 2.2.
Z = Root of R square+xl square.
Cos = R/Z.
                     (1)
                
            
                        Ryan said: 
                         
                        1 decade ago
                
                Z = sqrt(1500^2+2200^2) = 2663.
pf = R/Z = 1500/2663 = 0.564.
                pf = R/Z = 1500/2663 = 0.564.
                        Shenoy said: 
                         
                        1 decade ago
                
                Power factor always less than 1 use common sense, so simple.
                
                     (1)
                
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