Question 12.43: In Fig. 12.38, i(t)=I1cos(ω1t)+I2cos(ω2t+θ) , where ω1≠ω2....
In Fig. 12.38, \quad i(t)= I_1 \cos \left(\omega_1 t\right)+I_2 \cos \left(\omega_2 t+\theta\right) , where \omega_1 \neq \omega_2. Use superposition to obtain an expression for the voltage ν .

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Let \tilde{I}_1=I_1 \angle 0, \tilde{I}_2=I_2 \angle \theta and obtain expressions for the responses to these excitations individually, using the appropriate frequency in each case. Kirchhoff’s current law yields
\begin{aligned} &\frac{\tilde{V}}{R}+j \omega C \tilde{V}=\tilde{I} \\\\ &\Rightarrow \tilde{V}=\frac{R}{1+j \omega C R} \tilde{I} \end{aligned} ( 12.63)
The individual phasors for the individual responses are
\begin{aligned} &\tilde{V}_1=\frac{R}{1+j \omega_1 C R} \tilde{I}_1 \\\\ &\Rightarrow v_1(t)=\left|\tilde{V}_1\right| \cos \left(\omega_1 t+∡ \tilde{V}_1\right) \\\\ &\tilde{V}_2=\frac{R}{1+j \omega_2 C R} \tilde{I}_2 \\\\ &\Rightarrow v_2(t)=\left|\tilde{V}_2\right| \cos \left(\omega_2 t+\theta+∡ \tilde{V}_2\right) \end{aligned} (12.64)
By superposition,
\begin{aligned} ν(t)=& ν_1(t)+ν_2(t) \\\\ =&\left|\tilde{V}_1\right| \cos \left(\omega_1 t+\not \tilde{V}_1\right) \\\\ &+\left|\tilde{V}_2\right| \cos \left(\omega_2 t+\theta+∡ \tilde{V}_2\right) \end{aligned} ( 12.65)