99 Percentile Qs Bank for JEE MainPhysicsAlternating Current
An inductor and a resistor are connected in series to an AC source. The current in circuit is 500 ~mA , if the applied AC voltage is 8 2 ~V at a frequency of 175 Hz and the current in the circuit is 400 ~mA , if the same AC voltage at a frequency of 225 Hz is applied. The values of the inductance and the resistance are respectively
Options
- A60 mH , 71
- B60 mH , 71
- C60 mH , 71
- D60 mH , 71
Correct answer
D. 60 mH , 71
Step-by-step solution
aligned & For an L-R circuit, I= V Z = V R^2+L^2 ^2 & R^2+L^2 ^2= ( V I )^2 & Here, I₁=500 10⁻³ ~A & ₁= 175 2 rad s =350 rad s & V₁=8 2 & R^2+L^2(350)^2= ( 8 2 500 10⁻³ )^2 & aligned Solving, we get R= 71 and L=60 mH