Question

# A resistance of 20 Ω is connected to a source of alternating current rated 110 V,50 Hz. Find the time taken by the current to change from its first maximum value to the consecutive rms value.

A
5.5 ms
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4.5 ms
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2.5 ms
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
3.5 ms
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 2.5 msLet the current be, i=i0sinωt. Also, let t1 and t2 be the time instants for consecutive appearances of the maximum value and the rms value of the current, respectively. i0=i0sinωt1 ⇒ωt1=π2⇒t1=π2ω i0√2=i0sinωt2 ⇒ωt2=3π4⇒t2=3π4ω [As we are taking the rms value after the first peak value] So, the time taken by the current to change from its first maximum value to the consecutive rms value is, t2−t1=3π4ω−π2ω=π4ω ⇒t2−t1=π4(2πf)=18×50=2.5×10−3 s=2.5 ms

Suggest Corrections
15
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program