A coil has an inductance 0.7 H and it is joined in series with a resistance of 220 `Omega`. When AC of 220 V, 50 Hz is applied to it, then wattles component in the circuit is

Options

A. `5 A`
B. `0.5A`
C. `0.7A`
D. `7A`

Answer

B

Explanation

Given, inductance, L = 0.7 H, resistance R = 200 `Omega`
AC potential, `V_(0)=200V` and frequency, `f=50Hz`
According to the question, we draw the following circuit,

Since, in a series circuit, the current is same for all series connected elements.
So, `" "I_(0)=V/Z=220/sqrt(R^(2)+(omegaL)^(2))" "[because omega=2pif]`
`rArr " "I_(0)=220/sqrt((220)^(2)+(2pi times 50 times 0.7)^(2))`
`" "=220/sqrt(48400+48312.04)`
`" "=220/311=0.7A`
As current in an AC circuit is given as,
`I_(0)=I_(R)+jI_(L)=Icosphi+jIsinphi`
where, `I_(L)=Isinphi` is wattless current
So, `" "I_(L)=0.7 times (omegaL)/Z " "(because sinphi=(omegaL)/Z)`
`" "=0.7 times (70pi)/311=0.495=0.5A`
Hence, the wattless component of current is 0.5 A.