Question

a resistor and a capacitor in a series ac circuit
a resistor with resistance ( r ) and a capacitor with capacitance
( c ) are connected in series to an ac voltage source. the time-
dependent voltage across the capacitor is given by
( v _ { c } ( t ) = v _ { c _ { 0 } } sin omega t ).
part a
what is the amplitude ( i _ { 0 } ) of the total current ( i ( t ) ) in the circuit?
express your answer in terms of any or all of ( r, c, v _ { c _ { 0 } } ), and ( omega ).
view available hint(s)
( i _ { 0 } = )

Explanation:

Step1: Find the capacitive reactance

The capacitive reactance \(X_C=\frac{1}{\omega C}\)

Step2: Find the impedance of the RC - series circuit

The impedance of an RC - series circuit is \(Z = \sqrt{R^{2}+X_{C}^{2}}=\sqrt{R^{2}+\frac{1}{\omega^{2}C^{2}}}\)

Step3: Use Ohm's law for AC circuits

Ohm's law for AC circuits is \(I_{0}=\frac{V_{0}}{Z}\). Since the voltage across the capacitor \(V_{C0}\) and the current \(I\) in an RC - series circuit are related by \(V_{C0}=I_{0}X_{C}\), we can also use another approach.

We know that \(V_{C}(t)=V_{C0}\sin\omega t\), and the current in a capacitor \(I = C\frac{dV_{C}}{dt}\). Differentiating \(V_{C}(t)\) with respect to \(t\): \(\frac{dV_{C}(t)}{dt}=V_{C0}\omega\cos\omega t\). The amplitude of the current \(I_{0}=C V_{C0}\omega\)

But wait, we should use the relation for the entire RC - circuit. The voltage of the source \(V(t)\) (not given directly, but we can use the fact that in an RC - series circuit, the current is the same through \(R\) and \(C\)).

The correct formula using the impedance concept:

We know that \(I_{0}=\frac{V_{0}}{Z}\), and if we consider the fact that \(V_{C0}=I_{0}X_{C}\), then \(I_{0}=\frac{V_{C0}}{X_{C}}\). But this is wrong because \(V_{C0}\) is not the source voltage.

The correct way:

The current in an RC - series circuit. The current \(I(t)\) is given by \(I(t)=\frac{V(t)}{Z}\). The amplitude of the current \(I_{0}=\frac{V_{0}}{Z}\). But if we consider the fact that \(V_{C}(t)\) is given. We know that \(V_{C}(t) = V_{C0}\sin\omega t\), and \(I(t)=C\frac{dV_{C}(t)}{dt}\).

Differentiating \(V_{C}(t)\) with respect to \(t\):

\(\frac{dV_{C}(t)}{dt}=\omega V_{C0}\cos\omega t\), so \(I_{0}=\omega C V_{C0}\)

Answer:

\(I_{0}=\omega C V_{C0}\)