Explanation:
Given that,
Capacitor = 30μC
Resistor = 49.0Ω
Voltage = 30.0 V
Frequency = 60.0 Hz
We need to calculate the impedance
Using formula of impedance
[tex]Z=\sqrt{R^2+X_{c}^2}[/tex].....(I)
We need to calculate the value of [tex]X_{c}[/tex]
Using formula of [tex]X_{c}[/tex]
[tex]X_{c}=\dfrac{1}{2\pi f c}[/tex]
[tex]X_{c}=\dfrac{1}{2\times\pi\times60.0\times30\times10^{-6}}[/tex]
[tex]X_{c}=88.42\ \Omega[/tex]
Put the value of [tex]X_{c}[/tex] into the formula of impedance
[tex]Z=\sqrt{(49.0)^2+(88.42)^2}[/tex]
[tex]Z=101.08\ \Omega[/tex]
(a). We need to calculate the rms current in the circuit
Using formula of rms current
[tex]I_{rms}=\dfrac{V}{Z}[/tex]
[tex]I_{rms}=\dfrac{30.0}{101.08}[/tex]
[tex]I_{rms}=0.30\ A[/tex]
The rms current in the circuit is 0.30 A.
(b). We need to calculate the rms voltage drop across the resistor
Using formula of rms voltage
[tex]V_{rms}=I_{rms}\times R[/tex]
Put the value into the formula
[tex]V_{rms}=0.30\times49.0[/tex]
[tex]V_{rms}=14.7\ V[/tex]
The rms voltage drop across the resistor is 14.7 V
(c). We need to calculate the rms voltage drop across the capacitor
Using formula of rms voltage
[tex]V_{rms}=I_{rms}\times X_{C}[/tex]
[tex]V_{rms}=0.30\times88.42[/tex]
[tex]V_{rms}=26.53\ V[/tex]
The rms voltage drop across the capacitor is 26.53 V.
Hence, This is the required solution.