tyz is going to have a 18220 midterm next week so he is going to review some simple RLC circuits!
RC
Imagine there is a resistor and a capacitor connected in serial. Let $I_C$ be the current passing through the capacitor, $V_C$ be the voltage across the capacitor. Then the voltage across the resistor is $-V_C$. Assume the initial voltage across the capacitor is $V_0$.
Since $I_C = C{ {d V_C} \over {d t} }$, we have $R C{ {d V_C} \over {d t} } + V_C = 0$, and ${ {d V_C} \over {d t} } + {1 \over {RC} } V_C = 0$. Solving this equation we have $V_C = A e^{- {t \over {RC} } }$. Applying the initial value, we have $V_C = V_0 e^{- {t \over {RC} } }$. Let time constant $\tau = RC$, $V_C = V_0 e^{- {t \over \tau} }$
RC with voltage source
There is a resistor and a capacitor, and also a voltage source $V_0$ in serial. This time we assume initial voltage across the capacitor is 0V. As capacitor eventually become open circuit, we know the voltage across capacitor eventually becom $V_0$.
Since $I_C = C{ {d V_C} \over {d t} }$, we have $R C{ {d V_C} \over {d t} } + V_C = V_0$, or ${ {d V_C} \over {d t} } + { {V_C} \over {RC} } = { {V_0} \over {RC} }$. Solving this differential equation, we have $V_C = A e^{- {t \over {RC} } } + V0$. Applying initial value we have $V_C = A (1 - e^{- {t \over \tau} })$.
RL
Imagine there is a resisor and an inductor connected in serial. Let $I_L$ be the current passing through the inductor (and also the resistor), $V_L$ be the voltage across the inductor. Assume initial current passing the inductor is $I_0$.
Since $V_L = L { {d I_L} \over {dt} }$, we have $L { {d I_L} \over {dt} } + R I_L = 0$, or ${ {d I_L} \over {dt} } + {R \over L} I_L = 0$. Solving this we got $I_L = A e^{- {R \over L} t }$. Let $\tau = {L \over R}$ and applying initial value, we have $I_L = I_0 e^{- {t \over \tau} }$.
RL with voltage source
Now we have a voltage source $V_0$ in addition! As inductor eventually become short circuit, we know the current through the inductor eventually become $I_0$. Assume initial current through the inductor is 0A.
Since $V_L = L { {d I_L} \over {dt} }$, we have $L { {d I_L} \over {dt} } + R I_L = V_0$, ${ {d I_L} \over {dt} } + {R \over L} I_L = {V_0 \over L}$. So we have $I_L = {V_0 \over R} + A e^{- {t \over \tau} }$. Applying the initial value we have $I_L = {V_0 \over R} (1 - e^{- {t \over \tau} })$.
LC
Imagine there is a inductor and a capacitor in serial, and we want to know the voltage across the capacitor $V_C$. Then voltage across inductor is $-V_C$.
By $I = C{ {d V_C} \over {dt} }$ and $-V_C = L { {dI} \over {dt} }$, we have $-V_C = LC { {d^2 V_C} \over {dt^2} }$, or ${ {d^2 V_C} \over {dt^2} } + {V_C \over {LC} } = 0$. Let $\omega_0 = \sqrt{1 \over {LC} }$, we have $V_C = C_1 \cos (\omega_0 t) + C_2 \sin (\omega_0 t)$, and $V_C = R \sin (\omega_0 t + \phi)$.
RLC
Now suppose we have a resistor, a inductor and a capacitor in parallel! And we also want to know $V_C$.
Since $-V_C - IR = L { {dI} \over {dt} }$, we have ${ {d^2 V_C} \over {dt^2} } + {R \over L} { {d V_C} \over {dt} } + {V_C \over {LC} } = 0$. Let $\alpha = {R \over {2L} }$, solving $r^2 + 2\alpha r + (\omega_0)^2 = 0$. Now solution is (assume $\omega_0 > \alpha$), let $\mu = \sqrt{\omega_0 ^ 2 - \alpha ^ 2}$, $V_C = e ^ {-\alpha} (C_1 \cos (\mu t) + C_2 \sin (\mu t) )$, and $V_C = e ^ {-\alpha} R \sin (\mu t + \phi) $
Other circuits
For more complex circuits, we can always do nodal analysis! (QAQ)
- Label nodes and their voltages
- Write KCL formulas
- For voltage source, create a new variable of the current flowing through it and add an equation about the voltage difference across it
- For inductor, create a new variable of the current flowing through it and add an equation about the inductor voltage-current relation
