# A 200ohm R is connected to an a.c. source with max volt of 10V. what is the max I if freq is 100Hz and when R is replaced with 0.5H L and 250microF C?i'm not sure to use wave equation...

A 200ohm R is connected to an a.c. source with max volt of 10V. what is the max I if freq is 100Hz and when R is replaced with 0.5H L and 250microF C?

i'm not sure to use wave equation V=V(peak)cos(2PIft) or other equation.

### 1 Answer | Add Yours

The angular frequency of the voltage source is

`omega= 2*pi*F = 2*pi*100 =628.31 (rad)/s`

There are two different cases:

1. The resistor is replaced by a series combination of L =0.5 H and C=250 `muF` . In this case the total impedance is

`Z = X_L-X_C = omega*L -1/(omega*C) = 628.31*0.5 -1/(628.31*250*10^-6) =314.16-6.37=307.79 ohm`

The total current is given by the Ohm's law

`I = U/Z = 10*cos(omega*t)/307.79 =0.0325*cos(omega*t) A`

`I_(max) = 0.0325 A =32.5 mA`

2. The resistor is replaced by a parallel combination of L and C. In this case the total impedance is

`Z = (X_L*X_C)/(X_L +X_C) = ((omega*L)/(omega*C))/(omega*L +1/(omega*C)) =(314.16*6.37)/(314.16+6.37) =6.243 ohm`

The total current is

`I = U/Z = (10*cos(omega*t))/6.243 =1.602*cos(omega*t) A`

The maximum current in this case is

`I_(max) =1.602 A`