The voltage and current in an a.c. circuit are V = 50 sin 314 tV and I = 10 sin 314 t + π 4 A respectively. Calculate: i Wattless current

Given, V = 50 sin 314 t v I = 10 sin 314 + π 4 A I r m s = 10 2 ϕ = p h a s e c o n s tan t = π 4 Wattless current, I = I r m s sin ϕ I = 10 2 × 1 2 I = 5 A

The voltage and current in an a.c. circuit are V = 50 sin 314 tV and I = 10 sin 314 t + π 4 A respectively. Calculate: ii Root mean square voltage

Given, V = 50 sin 314 tV and I = 10 sin 314 t + π 4 A V m a x is the maximum value of emf in the given AC circuit V m a x = 50 V We know that, V r m s = V m a x 2 (Root mean square voltage) V r m s = 50 2 V r m s = 25 2 V

The voltage and current in an a.c. circuit are V = 50 sin 314 tV and I = 10 sin 314 t + π 4 A respectively. Calculate: ii Root mean square voltage

Given, V = 50 sin 314 tV and I = 10 sin 314 t + π 4 A V m a x is the maximum value of emf in the given AC circuit V m a x = 50 V We know that, V r m s = V m a x 2 (Root mean square voltage) V r m s = 50 2 V r m s = 25 2 V

The voltage and current in an a.c. circuit are V = 50 sin 314 tV and I = 10 sin 314 t + π 4 A respectively. Calculate: i Wattless current

Given, V = 50 sin 314 t v I = 10 sin 314 + π 4 A I r m s = 10 2 ϕ = p h a s e c o n s tan t = π 4 Wattless current, I = I r m s sin ϕ I = 10 2 × 1 2 I = 5 A

What does meant by half powerpoint frequencies. Obtain an expression for bandwidth in a LCR series circuit. Show half power point frequency in curve between alternating current and frequency.

Half power point frequency: The input frequency at which the power dissipation in LCR circuit becomes half of maximum power. Resonance condition: Here, ω 1 = ω 0 - ∆ ω ω 2 = ω 0 + ∆ ω Band width ω 2 - ω 1 = ( ω 0 + ∆ ω ) - ( ω 0 - ∆ ω ) ω 2 - ω 1 = 2 ∆ ω The ω 1 and ω 2 has frequency at which maximum current becomes 1 2 times of it's value V 0 = I 0 Z V 0 Z = I 0 2 I 0 R Z = I 0 2 V 0 = I 0 R 2 R = Z 2 R = R 2 + X C - X L 2 2 R 2 = R 2 + ( 1 ω C - ω L ) R 2 = 1 ω 1 C - ω 1 L 2 R = 1 ω 0 C 1 + ∆ ω ω 0 - ω 0 L 1 + ∆ ω ω 0 R = ω 0 L 1 + ∆ ω ω 0 - 1 + ω 0 L 1 + ∆ ω ω 0 R = ω 0 L 1 - ∆ ω ω 0 - ω 0 L 1 + ∆ ω ω 0 R = ω 0 L 1 - ∆ ω ω 0 - 1 - ∆ ω ω 0 R = ω 0 L 2 ∆ ω ω 0 R = L 2 ∆ ω ∆ ω = R 2 L B a n d w i d t h

Write one merit and one de-merit of alternating current in comparison of direct current. Obtain expression for following in a pure inductive alternating current circuit- iii Peak value of current Draw curve for power in pure inductive circuit.

Advantage The generation of AC is more economical than DC Disadvantage The peak value of AC is high. It is dangerous to work with AC( I 0 = 2 I ) Pure inductor V 0 sin ( ω t ) - L d i d t = 0 V 0 sin ( ω t ) = L d i d t Integrating both sides we get, ∫ V 0 L sin ( ω t ) d t = d i - V 0 L ω cos ( ω t ) = i - V 0 X L sin ( π 2 - ω t ) = i - i 0 sin ( π 2 - ω t ) = i i = i 0 sin ( ω t - π 2 ) (i) i 0 = V 0 X L = V 0 ω L ← p e a k value of current (ii) X L = ω L ← i n d u c t i v e r e a c tan c e (iii) i = i 0 sin ( ω t - π 2 ) ← i n s tan tan e o u s v a l u e