Alternating Current Part 1

Given, resistance of resistor =R
Inductance, XL=3R
Capacitance XC=2R
As we know that,
Power factor, cos‌φ=‌

R

Z

where, Z is impedance,
Z=√R2+(XL−XC)2
For case I, an inductance connected in series with resistance
cos‌φ1‌‌=‌

R

Z1

=‌

R

√R2+(XL−0)2

‌‌=‌

R

√R2+(3R)2

=‌

R

R√10

=‌

1

√10

‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
For case II, A capacitor is also connected in series with resistance
∴‌‌cos‌φ2‌‌=‌

R

Z2

⇒‌‌cos‌φ2‌‌=‌

R

√R2+(3R−2R)2

‌‌=‌

R

√R2+R2

=‌

R

R√2

=‌

1

√2

...‌ (ii) ‌
Now, ratio of Eqs. (ii) and (i), we get
‌‌‌

cos‌φ2

cos‌φ1

‌‌=‌

‌ 1 √2

1 √10

=√‌

10

2

=√5:1
∴‌‌x‌‌=1