CBSE Class 12 · 2007 · Alternating CurrentShort Answer

Given are two electric circuits A and B. In circuit A an inductor of reactance

X_{L} = 3 R

is in series with a resistance

R

. In circuit B an inductor of reactance

X_{L} = 3 R

, a resistance

R

and a capacitor of reactance

X_{C} = R

are all in series. Calculate the ratio of the power factor of circuit B to the power factor of circuit A.

Answer: $\frac{p.f. of B}{p.f. of A} = 2$

Solution

Power factor

= cos ϕ = \frac{R}{Z}

.
Circuit A (

R

and

X_{L} = 3 R

cos ϕ_{A} = \frac{R}{R ^{2} + ( 3 R ) ^{2}} = \frac{R}{10 R} = \frac{1}{10}

Circuit B (

R

X_{L} = 3 R

X_{C} = R

cos ϕ_{B} = \frac{R}{R ^{2} + ( 3 R - R ) ^{2}} = \frac{R}{R ^{2} + 4 R ^{2}} = \frac{1}{5}

Hence

\frac{cos ϕ _{B}}{cos ϕ _{A}} = \frac{1/ 5}{1/ 10} = \frac{10}{5} = 2

\frac{p.f. of B}{p.f. of A} = 2

CBSE Class 12 Board · 2007 · Subjective