Question

$\frac { 2.5 } { \pi } \mu F$ capacitor and 3000-ohm resistance are joined in series to an ac source of 200 volt and $50 \mathrm { sec } ^ { - 1 }$ frequency. The power factor of the circuit and the power dissipated in it will respectively

• A. 0.6, 0.06 W
• B. 0.06, 0.6 W
• C. 0.6, 4.8 W
• D. 4.8, 0.6 W

Answer 1

Answer: (C)

0.6, 4.8 W

Answer 2

$\mathrm { Z } = \sqrt { \mathrm { R } ^ { 2 } + \left( \frac { 1 } { 2 \pi \mathrm { vC } } \right) ^ { 2 } } = \sqrt { ( 1000 ) ^ { 2 } + \frac { 1 } { \left( 2 \pi \times 50 \times \frac { 2.5 } { \pi } \times 10 ^ { - 6 } \right) ^ { 2 } } }$

⇒ $Z = \sqrt { ( 3000 ) ^ { 2 } + ( 4000 ) ^ { 2 } } = 8 \times 10 ^ { 3 } \Omega$

So power factor $\cos \phi = \frac { R } { Z } = \frac { 3000 } { 5 \times 10 ^ { 3 } } = 0.6$ and power

$P = \frac { ( 200 ) ^ { 2 } \times 0.6 } { 5 \times 10 ^ { 3 } } = 4.8 \mathrm {~W}$