Question

For the circuit shown, R = 75.0 ohms, L= 55.0 mg, and C = 25.0 μC. The power source has 12.0 V arms and a frequency of 60.0 Hz. a. what is the value of XL
b. what is the value of XC
c. What is the value of Z
d. What is the rms current
e. what is the rms voltage across the resistor
f. what is the rms voltage across the inductors
g. what is the rms voltage across the capacitorh h. how much power is dissipated by the circuit

Answer 1

Explanation:

Given that,

Resistance R = 75.0 ohms

Inductance L = 55.0 mH

Capacitance $C = 25.0\ \mu C$

Voltage V = 12.0 V

Frequency f = 60.0 Hz

We need to calculate the angular frequency

Using formula of angular frequency

$\omega = 2\pi f$

Put the value into the formula

$\omega =2\times3.14\times60.0$

$\omega=376.8\ rad/s$

(a). We need to calculate the  value of $X_{L}$

Using formula of $X_{L}$

$X_{L}=\omega\times L$

Put the value into the formula

$X_{L}=376.8\times55.0\times10^{-3}$

$X_{L}=20.724\ \Omega$

(b). We need to calculate the  value of $X_{L}$

Using formula of $X_{C}$

$X_{C}=\dfrac{1}{\omega C}$

$X_{C}=\dfrac{1}{376.8\times25.0\times10^{-6}}$

$X_{C}=106.16\ \Omega$

(c). We need to calculate the value of Z

Using formula of impedance

$Z=\sqrt{R^2+(X_{L}-X_{C})^2}$

Put the value into the formula

$Z=\sqrt{75.0^2+(20.724-106.16)^2}$

$Z=113.68\ \Omega$

(d). We need to calculate the rms current

Firstly we need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

Put the value into the formula

$I=\dfrac{12.0}{75.0}$

$I=0.16\ A$

Using formula of rms current

$I_{rms}=\dfrac{I_{0}}{\sqrt{2}}$

$I_{rms}=\dfrac{0.16}{\sqrt{2}}$

$I_{rms}=0.113\ A$

(e). We need to calculate the rms voltage across the resistor

Using formula of rms voltage

$V_{rms}=I_{rms}\times R$

$V_{rms}=0.113\times75.0$

$V_{rms}=8.475\ V$

(f). We need to calculate the rms voltage across the inductor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{L}$

$V_{rms}=0.113\times20.724$

$V_{rms}=2.342\ V$

(g). We need to calculate the rms voltage across the capacitor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{C}$

$V_{rms}=0.113\times106.16$

$V_{rms}=11.99\ V$

(h).  We need to calculate the dissipated power by the circuit

Using formula of dissipated power

$P=RI^2$

Put the value into the formula

$P=75.0\times0.113^2$

$P=0.958\ W$

Hence, This is the required solution.