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+### Statement
+
+$11.4.12.$ [Insert the problem statement]
+
+### Solution
+
+a) The voltage drop across the resistor is just Ohm's law:
+
+$$V_R = I_0 R cos \omegat$$
+
+There is a charge on the capacitor:
+
+$$q = V_C C= \int_{0}^{t} I(t) dt$$
+
+$$V_C C = \frac{I_0}{\omega} sin(\omega t)$$
+
+$$V_C = \frac{I_0}{\omega C} cos(\omega t - \frac{\pi}{2})$$
+
+Finally, we write down the equation for the voltage across the coil:
+
+$$V_L = L \frac{dI}{dt} = -LI_0 \omega sin(\omegat)$$
+
+$$V_L = \frac{dI}{dt} = LI_0 \omega cos(\omegat + \frac{\pi}{2})$$
+
+The phase shift as seen in the coil $+\frac{\pi}{2}$ depends on the current, in the capacitor $-\frac{\pi}{2}$, the resistance has no shift, the amplitudes are equal:
+
+$$V_R = I_0 R$$
+$$V_C = \frac{I_0}{\omega C}$$
+$$V_L = LI_0 \omega$$
+
+b)
+To find the amplitude, it is enough to add up our voltages vector based on the diagram obtained in a:
+
+$$V = I_0 \sqrt{R^2 + (\omegaL - \frac{1}{\omega C})^2}$$
+
+Since we have a right triangle formed from vectors, the phase shift is easy to express in terms of a tangent.:
+
+$$\varphi = arctg \frac{\omegaL - \frac{1}{\omega C}{R}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]