WAEC Physics 2017 Objective — Question 43

Question 43 of 50 from the West African Examinations Council (WAEC) Physics 2017 Objective paper, with the correct answer and a full explanation.

The resonant frequency of an a.c. circuit is 1000 kHz. If each of the capacitance and inductance in the circuit is reduced by 50% and no other changes are made, the resonant frequency will become

A. A. 250 kHz
B. B. 750 kHz
C. C. 1000 kHz
D. D. 2000 kHzCorrect

Explanation

f0=1/(2π√LC). f0∝1/√LC. Given f1=1000kHz, L1=L, C1=C; L2=L/2, C2=C/2 (both reduced by 50%). 1000√(L1C1)=f2√(L2C2) => f2 = 1000√(L1C1)/√(L2C2) = 1000x√(LC)/√((L/2)(C/2)) = 1000x2=2000kHz.

Sign up free to unlock

Score tracking
Practice history
Saved questions
Progress dashboard
Personalized sessions
Weak-topic breakdown

…and/or go further with premium services and No Ads.

Create free account

More questions from Physics 2017 Objective More Physics past papers