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Malhar Rajpal In physics, a capacitor is a device that stores electrical energy. A capacitor is made up of two plates with a dielectric material between the plates. When a voltage is applied to a circuit with a capacitor, positive charge builds up on one plate while negative charge builds up on another plate, with the result being a potential difference (potential refers to the electrical potential as a result of the different charge between the plates). When the potential difference of the capacitor is the same as the voltage applied in the circuit, the current in the circuit becomes 0 and the capacitor’s p.d. stops increasing. The ‘charging’ process for the capacitor happens very quickly when there is no resistance in the circuit. If there is a resistance, by I=V/R, the current is lower and thus the charge builds up slower on the plates, slowing down the charging process. The more resistance, the slower the capacitor reaches the voltage applied. A capacitor also has a property called capacitance, C, which is the ratio of the amount of electric charge stored on a conductor to the difference in electric potential (this is directly proportional to the area of the plates and inversely proportional to the distance between the plates). The maximum charge on a capacitor is given by the equation Q = CV, where Q is charge, C is capacitance, and V is voltage. The capacitor can then be removed from the circuit with the power source and put in another circuit without a power supply. The charge then dissipates from the capacitor and flows through the circuit. This causes the absolute value of the charges on the plates and therefore the potential difference to decrease until they reach 0. The rate that the charge decreases with time of some capacitor with capacitance C depends on the resistance of the circuit and today we will explore the mathematical model finding the charge Q in this circuit at any given time while the charge dissipates. In the new loop, with the capacitor and some resistance, R, the current I of the circuit can be modelled by: I=V/R and V=IR. From above, we have the equation linking charge to capacitance and p.d.: Q = CV. Rearranging this equation, we see that V = Q/C. Substituting V = IR, IR = Q/C and thus I=Q/(RC). Current is also defined as the net rate of flow of electric charge, which can be defined mathematically as the change in charge over the change in time: ΔQ/Δt. Since charge is being dissipated (decreasing), we can say that ΔQ/Δt must be negative and, being physicists (this step is illegal is maths!), we can can say that I = -ΔQ/Δt. Now that we have two expressions for I, we can combine them: -ΔQ/Δt = Q/(RC), a differential equation! Using calculus notation, -dQ/dt=Q/(RC). Rearranging this equation, we get (1/Q)dQ = -1/(RC) dt. We can now integrate both sides: Since we don’t want the unknown constants resulting from indefinite integration in our final expression, we need to make both integrals definite (along a fixed range). We can say that we start at t=0 and go until t=t. When t=0 seconds, Q = Q_0 (initial capacitance), and when t=t seconds, we can say that Q=Q (some capacitance). So we get: Solving this, since -1/(RC) is a constant with respect to t, we can bring it out of the integral to get: The indefinite integral of 1/Q dQ is just ln(Q) + C (natural log of Q), where C is an arbitrary constant, so by using the definite integrals, the left side of the equation is: The indefinite integral of dt is just t+c and thus the right side becomes: Using ln(a) - ln(b) = ln(a/b), ln(Q) - ln(Q_0) = ln(Q/Q_0). We can now raise e (Euler’s constant) to the power of both sides to get: Since e^(ln(a)) = a for any a, we can rewrite as Q/Q_0= e^(-t/(RC)) and thus: where Q is charge, Q_0 is initial charge, e is Euler’s constant, t is time since discharge begun, R is resistance, and C is capacitance.
The units of RC is seconds and it is thus be referred to as the time constant, τ. Using Q=Q_0 e^(-t/(RC)), when t = τ, Q = Q_0/e; when t = 2τ, Q = Q_0/e^2; when t = 3τ, Q = Q_0/e^3. In every discharge cycle, the time constant is equal to the amount of time taken for the charge to reduce to ~37% (1/e) of its initial value. Since charge is easily measured and this time is a constant, one great use for capacitors is in timing circuits! As an exercise for yourself, try showing that the equations V=V_0 e^(-t/(RC)) and I = (V_0 / R) e^(-t/(RC)) also hold true in RC circuits!
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