consider a series R-C circuit consisting of a 30 ohm resistor and 20*10^-6F capacitor connected to a 100V,50Hz power supply.find the magnitude of the current and the angle by which the current leads the voltage.

Xc = 1/WC = -1/(314.2*20*10^-6) = -159.2 Ohms.

Z(Ohms) = 30 -j159.2 = 162[-79.3o].

I = E/Z = 100[0o]/162[-79.3] = 0.617A[79.3o].

To find the magnitude of the current and the phase angle by which the current leads the voltage in a series R-C circuit, we can use the concept of impedance and phase angle of a capacitor.

Impedance (Z) in a series R-C circuit is given by the formula:

Z = √(R^2 + (1 / (ωC))^2)

where:
- R is the resistance in the circuit (30 ohms in this case)
- C is the capacitance in Farads (20*10^-6 F in this case)
- ω is the angular frequency, which is equal to 2πf (f is the frequency in Hz)

Since the circuit is connected to a 100V, 50Hz power supply, we can calculate ω as follows:

ω = 2πf = 2π * 50 = 100π rad/s

Substituting the values into the impedance formula:

Z = √(30^2 + (1 / (100π * 20*10^-6))^2)
Z = √(900 + (1 / (200π))^2)
Z ≈ √(900 + 0.000401)^2)
Z ≈ √(900 + 0.000001)
Z ≈ √900.000001
Z ≈ 30.000005 ohms

The magnitude of the current (I) in the circuit can be found by dividing the supply voltage (V) by the impedance (Z):

I = V / Z
I = 100V / 30.000005 ohms
I ≈ 3.333 A

To find the phase angle by which the current leads the voltage, we can use the formula:

θ = arctan(-1 / (ωRC))

Substituting the values:

θ = arctan(-1 / (100π * 20*10^-6 * 30))
θ ≈ arctan(-1 / (600π*10^-6))
θ ≈ arctan(-1 / (0.00188682))
θ ≈ arctan(-530.303)
θ ≈ -89.948 degrees

Therefore, the magnitude of the current is approximately 3.333 A and the current leads the voltage by an angle of approximately 89.948 degrees.