Circuit theory finds various applications in circuit related problems == Both Voltage and Current == Initial conditions describe the energy stored in every capacitor and every inductor. Initial conditions are completely specified only when both voltage and current for all capacitors and all inductors is known. == Assume Zero == Assume zero for the capacitor voltage and/or inductor current if no information is given. == Inductor Initial Conditions == Every inductor has two initial conditions: current and voltage. When a switch is thrown that eliminates all power supplies, (or connects new power supplies) the inductor can turn into a power supply itself. The current through an inductor maintains it's direction and magnitude between t ( 0 − ) {\displaystyle t(0_{-})} and t ( 0 + ) {\displaystyle t(0_{+})} . The voltage may instantaneously switch polarity and/or magnitude. Energy changes elsewhere in a circuit cause an inductor to block these changes by appearing as an open. An Inductors will allow any voltage to appear across it's terminals in order to maintain the energy in its magnetic field. An inductor doesn't want the current to increase or decrease. Any current change will change the energy stored in its magnetic field. But gradually, an inductor will accept change. And after a very long time it acts like a wire or short allowing current only limited by other devices in the circuit to pass through it. == Capacitor Initial Conditions == Every capacitor has two initial conditions: voltage and current. When a switch is thrown that eliminates all power supplies, (or adds power) the capacitor can turn into a power supply itself. A capacitor maintains it's voltage polarity and value between t ( 0 − ) {\displaystyle t(0_{-})} and t ( 0 + ) {\displaystyle t(0_{+})} , but the current may instantaneously switch directions and may switch magnitude. A capacitor reacts to energy changes by instantly turning itself into a short or an ideal wire with no resistance. A capacitor doesn't care about the current flowing through it. A capacitor doesn't want the energy stored in the electric field between its plates to change. So the voltage is maintained, but the current can switch direction and magnitude. The current will only be limited by other devices in the circuit. But after a very long time, a capacitor gradually adapts to change and eventually develops a voltage that the rest of the circuit doesn't try to change. At this point the capacitor turns into an open. Inductors and capacitors both react to energy changes, but in exactly the opposite ways. == Particular Solution == The particular solution in response to a step function turns into a set of conditions that have to be present at t = ∞. These conditions should be true at all times .. and can be used to help evaluate constants. == Progression == Start with an expression with a voltage across the capacitor. This way the progression through all possible initial conditions requires taking derivatives: H ( s ) = V C V s ⇒ V C ( t ) ⇒ i ( t ) = C d V C ( t ) d t ⇒ V L = L d i ( t ) d t {\displaystyle H(s)={\frac {\mathbb {V} _{C}}{\mathbb {V} _{s}}}\Rightarrow V_{C}(t)\Rightarrow i(t)=C{dV_{C}(t) \over dt}\Rightarrow V_{L}=L{di(t) \over dt}} Starting at V_L means evaluating integrals and their constants: H ( s ) = V L V s ⇒ V L ( t ) ⇒ i ( t ) = 1 L ∫ V L d t ⇒ V C = 1 C ∫ i ( t ) d t {\displaystyle H(s)={\frac {\mathbb {V} _{L}}{\mathbb {V} _{s}}}\Rightarrow V_{L}(t)\Rightarrow i(t)={\frac {1}{L}}\int V_{L}dt\Rightarrow V_{C}={\frac {1}{C}}\int i(t)dt} Starting at the current means that may have to do both a derivative and an integral to evaluate all constants. This detail is displayed by doing one example problem four different ways. == Summary ==