Direct Current Circuits February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 1 Ammeters and Voltmeters ! A device used to measure current is called an ammeter ! A device used to measure potential difference is called a voltmeter ! To measure the current, the ammeter must be placed in the circuit in series Voltmeter in parallel High resistance ! To measure the potential difference, the voltmeter must be wired in parallel with the component across which the potential difference is to be measured February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 Ammeter in series Low resistance 2 Real-life Voltmeter in a Simple Circuit ! Consider a simple circuit consisting of a source of emf with potential difference Vemf = 150. V and a resistor with resistance R = 100 kΩ. ! A voltmeter with resistance RV = 10.0 MΩ is connected across the resistor. PROBLEM 1: ! What is the current in the circuit before the voltmeter is connected? SOLUTION 1: ! Ohm’s Law allows us to calculate the current before: Vemf 150. V −3 i= = = 1.50⋅10 A = 1.50 mA 3 R 100.⋅10 Ω February 19, 2014 Chapter 26 6 Real-life Voltmeter in a Simple Circuit PROBLEM 2: ! What is the current in the circuit after the voltmeter is connected? SOLUTION 2: ! The equivalent resistance of the resistor and the voltmeter connected in parallel is given by: 1 1 1 = + Req R RV ! The equivalent resistance is: 3 6 100.⋅10 Ω 10.0⋅10 Ω) ( )( RRV 4 Req = = = 9.90⋅10 Ω = 99.0 kΩ 3 6 R + RV 100.⋅10 Ω +10.0⋅10 Ω February 19, 2014 Chapter 26 7 Real-life Voltmeter in a Simple Circuit ! The current after connecting the voltmeter is: Vemf 150. V −3 i= = = 1.52⋅10 A = 1.52 mA 4 R 99.0⋅10 Ω ! So the current in the circuit increases by 0.02 mA, when the voltmeter is connected because the parallel combination of the resistor and the voltmeter has a lower resistance than that of the resistor alone. ! The effect is small. ! The larger the resistance of the voltmeter, the better. February 19, 2014 Chapter 26 8 RC Circuits ! So far we have dealt with circuits containing sources of emf and resistors ! The currents in these circuits did not vary in time ! Now we will study circuits that contain capacitors as well as sources of emf and resistors ! These RC circuits have currents that vary with time ! Consider a circuit with • a source of emf, Vemf, • a resistor R, • a capacitor C February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 9 Charging a Capacitor ! We start with the capacitor uncharged ! We close the switch and current begins to flow, building up opposite charges on the plates of the capacitor and creating a potential difference across the capacitor ! When the capacitor is fully charged, no more current flows ! The potential difference across the plates is equal to the potential difference provided by the source of emf ! The magnitude of the total charge on each plate is qtot = CVemf February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 10 Charging a Capacitor ! While the capacitor is charging, we can analyze the the current flowing by applying Kirchhoff’s Loop Rule q (t ) Vemf −VR −VC = Vemf − i (t ) R − =0 C ! The change of the charge on the capacitor is the current The term Vc is negative since dq (t ) the top plate of the capacitor i (t ) = is connected to the positive dt ! We can rewrite our equation as dq (t ) q (t ) R + = Vemf or higher potential - terminal of the battery. Thus analyzing counter-clockwise leads to a drop in voltage across the capacitor! dt C dq (t ) q (t ) Vemf dq + = Differential equation relating q and dt RC R dt February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 11 Charging a Capacitor ! The solution for this differential equation is t − ⎞ ⎛ q (t ) = qmax ⎜ 1− e τ ⎟ ⎝ ⎠ ! The constant τ is the time constant given by τ = RC ! The constant qmax is given by qmax = CVemf ! So we can write the solution to our differential equation t − ⎛ ⎞ RC q (t ) = CVemf ⎜ 1− e ⎟ ⎝ ⎠ Math Reminder: February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 12 Charging a Capacitor ! At t = 0 we have 0 − ⎛ ⎞ RC q ( 0 ) = CVemf ⎜ 1− e ⎟ = 0 ⎝ ⎠ ! At t = ∞ we have ∞ − ⎛ ⎞ RC q ( ∞ ) = CVemf ⎜ 1− e ⎟ = CVemf ⎝ ⎠ ! The current flowing through the circuit is the time derivative of the charge dq(t ) ⎛ Vemf ⎞ − RCt i (t ) = =⎜ ⎟⎠ e ⎝ dt R ! At t = 0 we have i (0) = Vemf R ! At t = ∞ we have ∞ ⎛ Vemf ⎞ − RC i (∞) = ⎜ e =0 ⎟ ⎝ ⎠ February 19, 2014 R Physics for Scientists & Engineers 2, Chapter 26 13 Charging a Capacitor February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 14 Discharging a Capacitor ! Now we consider a circuit containing only one resistor and a fully charged capacitor ! The charge on the capacitor is qmax ! Now we short the resistor across the capacitor by moving the switch from position 1 to position 2 ! Current flows until the capacitor is completely discharged ! While the capacitor is discharging, Kirchhoff’s Loop Rule gives us q (t ) −i (t ) R −VC = −i (t ) R − =0 C February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 15 Discharging a Capacitor ! Using the definition of current, we can rewrite our equation as dq(t ) q(t ) R dt + =0 C ! The solution to this differential equation is q (t ) = qmax e − t RC ! At t = 0 we have q ( 0 ) = qmax e − 0 RC = qmax ! At t = ∞ we have q ( ∞ ) = qmax e February 19, 2014 − ∞ RC =0 Physics for Scientists & Engineers 2, Chapter 26 16 Discharging a Capacitor ! The current flowing through the circuit is the time derivative of the charge qmax ) − RCt ( dq (t ) i (t ) = =− e dt RC ! At t = 0 we have qmax ) − RC0 qmax ) ( ( i (0) = − e =− RC RC ! At t = ∞ we have ∞ qmax ) − RC ( i (t ) = − e =0 RC February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 17 Discharging a Capacitor February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 18 Time Required to charge a Capacitor ! Consider a circuit consisting of a 12.0 V battery, a 50.0 Ω resistor, and a 100.0 μF capacitor wired in series ! The capacitor is initially completely discharged ! PROBLEM ! How long after the circuit is closed will it take to charge the capacitor to 90% of its maximum charge? SOLUTION ! The charge on the capacitor as a function of time is given by t − ⎛ ⎞ RC q (t ) = qmax ⎜ 1− e ⎟ ⎝ ⎠ February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 20 Time Required to charge a Capacitor ! We want to know the time until q(t)/qmax = 0.90 t t − − ⎛ ⎞ q t ( ) = 0.90 ⇒ 0.10=e RC RC 1− e = ⎜⎝ ⎟⎠ q max ! Taking the natural log of both sides gives us ⎛ − RCt ⎞ ln ( 0.10 ) =ln ⎜ e ⎟ ⎝ ⎠ t ⇒ ln ( 0.10 ) = − RC ! So the time to discharge the capacitor to 90% of its maximum charge is t = −RC ln ( 0.10 ) = − (50.0 Ω )(100.0⋅10−6 F )( −2.30 ) t = 0.0115 s = 11.5 ms February 19, 2014 Math Reminder: ln(ex)=x Physics for Scientists & Engineers 2, Chapter 26 21 Summary: RC Circuit ! Charging a capacitor: t − ⎞ ⎛ q(t ) = q0 ⎜1 − e τ ⎟ ⎝ ⎠ dq ⎛ Vemf i= =⎜ dt ⎝ R ⎞ ⎟e ⎠ ⎛ t ⎞ ⎜− ⎟ ⎝ RC ⎠ ! Discharging a capacitor: q = q0 e ⎛ t ⎞ ⎜− ⎟ ⎝ RC ⎠ dq ⎛ q0 ⎞ i= =⎜ ⎟e dt ⎝ RC ⎠ ! Time constant: ⎛ t ⎞ ⎜− ⎟ ⎝ RC ⎠ τ = RC Physics for Scientists & Engineers 2 22 Rate of Energy Storage in a Capacitor ! A resistor with R = 2.50 MΩ and a capacitor with C = 1.25 μF are connected in series with a battery for which Vemf = 12.0 V ! At t = 2.50 s after the circuit is closed, what is the rate at which energy is being stored in the capacitor? SOLUTION THINK ! When the circuit is closed, the capacitor begins to charge ! The rate at which energy is stored in the capacitor is given by the time derivative of the amount of energy stored in the capacitor, which is a function of the charge on the capacitor February 19, 2014 Physics for Scientists & Engineers 2, Chapter 25 23 Rate of Energy Storage in a Capacitor SKETCH ! The circuit contains a battery, a resistor, and a capacitor in series. February 19, 2014 Physics for Scientists & Engineers 2, Chapter 25 24 Rate of Energy Storage in a Capacitor RESEARCH ! The charge on the capacitor as a function of time is t − ⎛ ⎞ RC q (t ) = CVemf ⎜ 1− e ⎟ ⎝ ⎠ ! The energy stored in a capacitor with charge q is 1 q2 U= 2C ! The time derivative of the stored energy is dU d ⎛ 1 q 2 ⎞ q (t ) dq (t ) = ⎜ = ⎟ dt dt ⎝ 2 C ⎠ C dt ! The time derivative of the charge is the current dq (t ) ⎛ Vemf ⎞ − RCt i (t ) = =⎜ e ⎟ ⎝ R ⎠ dt February 19, 2014 Physics for Scientists & Engineers 2, Chapter 25 25 Rate of Energy Storage in a Capacitor SIMPLIFY ! Combining our equations t − ⎛ ⎞ RC CVemf ⎜ 1− e ⎟ ⎝ ⎠ ⎛ Vemf ⎞ − RCt dU q (t ) = i (t ) = ⎜⎝ ⎟⎠ e dt C C R t t 2 − − ⎛ ⎞ dU Vemf RC RC = e ⎜ 1− e ⎟ dt R ⎝ ⎠ CALCULATE ! We first calculate the time constant RC = ( 2.50⋅106 Ω )(1.25⋅10−6 F ) = 3.125 s February 19, 2014 Physics for Scientists & Engineers 2, Chapter 25 26 Rate of Energy Storage in a Capacitor ! Now we calculate the rate of change of energy stored in the capacitor 2.50 s 2.50 s − ⎛ ⎞ 12.0 V ) − 3.125 dU ( −5 s 3.125 s = e 1 − e = 1.425 ⋅10 W 6 ⎜ ⎟ dt 2.50 ⋅10 Ω ⎝ ⎠ 2 DOUBLE-CHECK (Energy Conservation) ! The current at t = 2.50 is 2.50 s − 12.0 V ⎛ ⎞ 3.125 s −6 i (t ) = ⎜ e = 2.16⋅10 A ⎟ 6 ⎝ 2.50⋅10 Ω ⎠ February 19, 2014 Physics for Scientists & Engineers 2, Chapter 25 27 Rate of Energy Storage in a Capacitor ! The rate of energy dissipation at t = 2.50 is 2 dU 2 −6 P= = i R = ( 2.16 ⋅10 A ) ( 2.50 ⋅106 Ω ) = 1.16 ⋅10−5 W dt ! The rate at which the battery delivers energy to the circuit at t = 2.50 s is P= dU = iVemf = ( 2.16 ⋅10−6 A ) (12.0 V ) = 2.59 ⋅10−5 W dt ! The energy supplied by the battery is equal to the sum of the energy dissipated in the resistor plus the energy stored in the capacitor February 19, 2014 Physics for Scientists & Engineers 2, Chapter 25 28