Edexcel A Level Physics:复习笔记7.13 Exponential Discharge in a Capacitor

The Discharge Equation

• When a capacitor discharges through a resistor, the charge stored on it decreases exponentially
• The amount of charge remaining on the capacitor Q after some elapsed time t is governed by the exponential decay equation:

• 
•  Where:
  • Q = charge remaining (C)
  • Q0 = initial charge stored (C)
  • e = exponential function
  • t = elapsed time (s)
  • R = circuit resistance (Ω)
  • C = capacitance (F)

Discharge Equation for Potential Difference

• The exponential decay equation for charge can be used to derive a decay equation for potential difference
• Recall the equation for charge Q = CV
  • It also follows that the initial charge Q0 = CV0 (where V0 is the initial potential difference)

• Therefore, substituting CV for Q into the original exponential decay equation gives:

• 
• Cancelling C from both sides gives the exponential decay equation for potential difference V:

• 
•  Where:
  • V = potential difference after some time t (V)
  • V0 = initial potential difference (V)
  • t = elapsed time (s)
  • R = resistance (Ω)
  • C = capacitance (F)
• This equation shows that the potential difference also decreases exponentially, from some initial value V0

• ---

Discharge Equation for Current

• The exponential decay equation for potential difference can be used to derive a decay equation for current
• Recall Ohm's law V = IR
  • It follows that the initial potential difference V0 = I0R (where I0 is the initial current)

• Therefore, substituting IR for V into the decay equation for potential difference gives:

• 
• Cancelling R from both sides gives the exponential decay equation for current I:

• 
•  Where:
  • I = current after some time t (A)
  • I0 = initial current (A)
  • t = elapsed time (s)
  • R = resistance (Ω)
  • C = capacitance (F)

• This equation shows that the current also decreases exponentially, from some initial value I0

Step 1: Write the known quantities

• • Initial potential difference V0 = 12 V
  • Resistance R = 1 kΩ = 1000 Ω
  • Capacitance C = 10 mF = 0.01 F
  • Time elapsed = 15 s

Step 2: Determine the initial current I0

• • Since the initial potential difference is 12 V and the resistance is 1000 Ω, then:
• 

Step 3: Write the decay equation for current

• • The decay equation for current is:

Step 4: Substitute quantities and calculate the current after 15 s

• • Substituting quantities gives the following:

I = (0.012) × (e–(15/(1000 × 0.01))

I = (0.012) × (e–1.5)

I = (0.012) × (0.223...)

I = 2.7 × 10–3 A = 2.7 mA

Natural Logarithms & Discharge Equations

• The exponential decay equations are not linear
• They can be turned into linear equations by using the natural logarithm function
• Recall the exponential decay equation for charge:

• Dividing both sides by Q0 gives:

• Taking the natural logarithm of both sides 'cancels' the exponential function e, giving:

• This simplifies to:

• Leaving an equation for the natural logarithm of charge Q as:

• This is the equation of a straight line graph, where:
  • ln Q is plotted on the y-axis
  • t is plotted on the x-axis
  • The gradient of the line is therefore equal to –1/RC

The natural logarithm of the exponential decay curve line arises it to a straight-line graph with a gradient equal to –1/RC

• Following similar steps, the linearised versions of the decay equations for potential difference V is:

• And for current I is:

Step 1: Write the equation for the linearised decay equation for potential difference

• • The linearised decay equation for potential difference is given by:

Step 2: Interpret the graph given using the linearised equation

• • The equation says the y-intercept of the straight line is represented by ln V0

Step 3: Use the y-intercept to determine the initial voltage

• • The y-intercept is equal to 2.1
  • Therefore:

ln V0 = 2.1

Step 4: Cancel the natural logarithm using the exponential function:

• • Raising both sides using the exponential function e 'cancels' the natural logarithm
  • This gives:

V0 = e(2.1) = 8.2 V