Question

A resistor (R 5 9.00 3 102 V), a capacitor (C 5 0.250 mF), and an inductor (L 5 2.50 H) are connected in series across a 2.40 3 102-Hz AC source for which DVmax 5 1.40 3 102 V. Calculate (a) the impedance of the circuit, (b) the maximum current delivered by the source, and (c) the phase angle between the current and voltage. (d)

Answer 1

Step-by-step explanation:

It is given that,

Resistance,
`R=9* 10^2\ \Omega`

Capacitance,
`C=0.25\ mF=0.25* 10^(-3)\ F`

Inductance, L = 2.5 H

Frequency,
`f=2.4* 10^2\ Hz`

Maximum voltage,
`V_(max)=1.4* 10^2\ V`

(a) Impedance of the circuit is given by :

`Z=√(R^2+(X_L-X_C)^2)`...............(1)

`X_L=2\pi fL`

`X_L=2\pi * 2.4* 10^2* 2.5=3769.91\ \Omega`

`X_C=(1)/(2\pi fC)`

`X_C=(1)/(2\pi * 2.4* 10^2* 0.25* 10^(-3))`

`X_C=2.65\ \Omega`

Impedance,
`Z=√((9* 10^2)^2+(3769.91-2.65)^2)`

Z = 3873.27 ohms

(b) Using Ohm's law,
`I_(max)=(V_(max))/(R)`

`I_(max)=(1.4* 10^2)/(9* 10^2)`

`I_(max)=0.15\ A`

(c) Phase angle is given by :

`\phi=tan^(-1)((X_L-X_C)/(R))`

`\phi=tan^(-1)((3769.91-2.65)/(9* 10^2))`

`\phi=76.5^(\circ)`

Hence, this is the required solution.