A 12V resistance and an inductance of `0.05/pi H`are connected in series. Across the end of this circuit an alternating voltage of 130 V and frequency 50 Hz is connected. Calculate the current in the circuit and the potential differnece across the inductance.

Explanation

`R= 10Omega`
`C = 0.1 muF = 0.1 xx10^(-6)F`
` E_v = 100 V`
`v = 50 Hz`
`I_v=?`
Impedance `= Z = sqrt(R^2+X_C^2)`
`=sqrt(R^2+(1/(2pivC))^2)`
`Z = sqrt((10)^(2) +1/((2xx3.14 xx50xx0.1xx10^(-6))^(2))`
`= sqrt(100+1/((314xx10^(-7))^(2)))`
`= 100 +(10^7 xx10^7)/(314 xx314)`
`=sqrt(100+31847.133xx31847.133)`
`=sqrt(100+10.14xx10^8)`
`= sqrt(10^8 [1+10.14xx10^6])`
`=10sqrt(10140001)=10xx3184.3`
`= 3.18 xx10^4Omega`
`I_v=E_v/Z`
`=100/(3.18 xx10^4)`
`= 0.314 xx10^(-2)`
`I_v = 3.14 xx10^(-3)A`
`= 3.14mA`