An alternating current having peak value 14A is used to heat a metal wire. To produce the same heating effect, a constant current I can be used. Then, the value of I must be
A
14A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
about 20A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
about 10A
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
7A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C about 10A Irms=Peakvalue√2 =14√2=9.9A