Answer:
113.0 Hz
Explanation:
In a RC circuit with alternating current, the voltage across the resistor is given by Ohm's Law:
[tex]V_R = IR[/tex]
where
I is the current in the circuit
R is the resistance of the resistor
While the voltage across the capacitor is given by:
[tex]V_C = IX_C[/tex]
where
[tex]X_C=\frac{1}{2\pi f C}[/tex] is the impedance of the capacitor, where
f is the frequency
C is the capacitance
The voltages across the resistor and the capacitor are equal when
[tex]V_R = V_C[/tex]
So
[tex]IR=\frac{I}{2\pi fC}[/tex]
Which can be rewritten as
[tex]f=\frac{1}{2\pi RC}[/tex]
In this problem:
[tex]R=59.4\Omega[/tex] is the resistance
[tex]C=23.7\mu F = 23.7\cdot 10^{-6}F[/tex] is the capacitance
Substituting, we find the frequency at which this happens:
[tex]f=\frac{1}{2\pi (59.4)(23.7\cdot 10^{-6})}=113.0 Hz[/tex]
The value of fd represents the amplitudes of the voltages across the resistor and capacitor equal to 113.0 Hz.
Here
V = IR
Here.
I is the current in the circuit
R is the resistance of the resistor
Now the voltage across the capacitor and resistor should be provided by
Vr = Vc
IR = 1/2πFc
Here
f = 1 /2πRC
So,
= 1/2π(59.4)(23.7.10^-60
= 113.0 Hz
Hence, The value of fd represents the amplitudes of the voltages across the resistor and capacitor equal to 113.0 Hz.
Learn more about ohm law here: https://brainly.com/question/17139885?referrer=searchResults
