feat. Oliver, Evan, Michelle, and John
Material property ρ Describes the microscopic structure of a conductor ▪ The stuff that an electron has to move through Total resistance depends on resistance What makes it harder for an electron to move? ▪ Longer wire ▪ Less cross-sectional area R=ρL/A
Deals with potential difference, resistance, and current Current (speed which charges can move) is directly related to how much they want to move (voltage) and inversely related to how much stuff is in their way (resistance) I=V/R V=IR
Kirchhoff’s Voltage Law (KVL) (Loops) Around a closed loop, the sum of all voltage changes must equal zero Kirchhoff’s Current Rule (KCL) (Nodes) The sum of currents flowing into a node equals the currents flowing out of a node
In series, current through each resistor is equal (no node splits), but voltage is split up, so… ΔV TOT = ΔV 1 + ΔV 2 IR TOT =IR 1 +IR 2 R TOT =R 1 +R 2
In parallel, voltage drop is the same (closed loops all over), but current is split up, so… I TOT =I 1 +I 2 V/ R T =V/ R 1 + V/ R 2 1/R T =1/R 1 +1/R 2
P= ΔW/ΔtΔW=(Δq)VI=Δq/t P=IV Using V=IR, we can substitute to get: P=IV=(V/R)(V)=V 2 /R P=IV=(I)(IR)=I 2 R
Math is on the handout.
Remember C=Q/V Capacitors in Parallel follow KVL C T =C 1 +C 2 Capacitors in Series follow KCL 1/C T =1/C 1 +1/C 2
Remember I=dQ/dt; C=Q/ ΔV Apply KVL: Over time, the amount of current in the circuit decreases
τ=RC Time constant for charging depends on resistance in the circuit and how much the capacitor can hold At 3τ, the device is 95% charged ▪ Steady state conditions Meanwhile while charging… I asymptotically approaches 0 as I = Q stored asymptotically increases to CV battery V cap approaches V battery
Rearranging gives you Constants depend on charge in circuit at t=0 and t=infinity But what really matters: At first, there is no charge in a capacitor. Over a long period of time, the charge equals the product of capacitance and battery voltage
From previous capacitor info U=1/2(QV)=1/2(C V 2 )=1/2( Q 2 /C)
Current in the circuit is defined as: V here is the V of the cap, not necessarily the V of supply Charge in the Capacitor is defined as: Q0 is the charge on the capacitor when the switch is flipped
A capacitor of capacitance C is discharging through a resistor of resistance R. Leave answers in terms of τ. When will the charge on the capacitor be half of its initial value? When will the energy stored in the capacitor be half of its initial value?
Half charge? Using discharging formula Adjust for half Take natural log Rearrange for t
Half energy? Using energy formula Adjust for discharging Adjust for half Take logs Rearrange for t U= 1/2( Q 2 /C)
So in case you haven’t been paying attention, what is an RL circuit?: As the name would imply it is a circuit with a resistor and inductor. ▪ What is a resistor?: It resists things. In the case of circuits it is resisting the flow of electrons. ▪ What is an inductor?: A circuit component that initially tries to oppose changes in the current running through it. After a while it stops fighting the change and begins acting like a wire.
So in an RL circuit that you just turn on the current in the resistor does not rise quickly to a value of ε/R, but it doesn’t. Why? ▪ The inductor in the circuit creates a self induced emf to oppose the rise of the current, which in turn means that the emf of the inductor is opposite in polarity to that of the battery. ▪ As long as the inductor is opposing the emf of the battery the resistor will have a value smaller than ε/R. As time goes on the discrepancy between what the current at the resistor should be and what it actually is will get smaller.
But Evan, what kind of formulas could we use in solving RL circuits? Well, I’m glad you asked: Loop rule for RL circuits.
But what if I want more forumals?:
A circuit consisting solely of an inductor and a capacitor In the circuit, stored energy oscillates at a specific resonant frequency