The figure below depicts a source and a resistance in series with an ammeter.

• The ammeter, which is the device at the top of the figure, consists of a parallel combination of a small resistance, and a large resistance in series with a coil and a magnetic field.
• Almost all of the current through the ammeter will travel through the small resistance, and because of the large resistance a very small current will travel through the coil.
• The current through the coil will be in a known ratio to the current through small resistance, so the deflection of the coil can be calibrated to accurately measure the current in the circuit.

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A voltmeter consists of a coil and a magnetic field, in series with a large resistance.

• When a voltmeter is connected in parallel with a capacitor, a voltage across a combination results initially in a relatively large current through the capacitor and, because of the large resistance of the voltmeter, a small current through the meter.
• As the capacitor builds charge, it develops a voltage opposing that of the source and the current through the capacitor decreases.
• Since the rate at which the capacitor builds charge is equal to the current, as the current decreases charge builds less and less quickly.

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When the capacitor has been charged, the source can be disconnected, leaving the capacitor in series with the voltmeter.

• The voltage across the capacitor will then cause current to flows through the voltmeter, and the resulting deflection of the coil will register the voltage across the capacitor.
• The capacitance C of the capacitor is its charge-to-voltage ratio C = Q / V.
• The large capacitance means that a lot of charge can be stored on the capacitor for given voltage.
• The voltage across the capacitor will therefore be V = Q / C. The current flowing through the meter carries electrons from the negative plate of the capacitor around the circuit to the positive, decreasing the negative charge in utilizing in equal amount of positive charge so that Q decreases.
• As the charge Q decreases, the voltage across the capacitor decreases, as does the current through the meter. The meter therefore registers a decreasing voltage.

In the figure below we see that the current I = V / R must be equal to Q / (R C).

• Since the current I represents the rate at which charge is decreasing on the capacitor, we have I = dQ / dT.
• Since current is decreasing we write dQ / dt = - Q / (R C).
• (The next few steps require the use of some basic calculus and may be ignored by non-calculus students).
• We solve this equation by placing all factors involving Q on the left and all factors involving t on the right side of the equation, as in the last line.
• As in the second figure below, we integrate both sides of this equation to obtain ln Q = -t / (R C) + c, where c is the integration constant.
• Exponentiating both sides we obtain the solution in the last line.

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(Non-calculus students can now tune back in).

• Since e^c is a positive constant for constant c, we can call this constant C and write Q = C e^(-t / (R C) ).
• This exponential function gives us the charge on the discharging capacitor as a function of clock time t.

We observe that a .47 Farad capacitor discharges in such away that an initial voltage of 5.17 V reduces after 20 minutes to 4.9 V.

• Since voltage is proportional to the charge of the capacitor, we have V(t) = C e^-(t / (R C)), where the constant C is proportional to the constant C of the Q(t) function.
• Since V(0) = C e^0 = 5.17 V we see that C = 5.17 V, so now V = 5.17 V e^-(t / (R C) ).
• Substituting t = 20 into this equation we see that V(20) = 5.17 V e^-(20 / ( R C ) ) = 4.9 V. As shown below we solve this equation for R C, obtaining R C = 400 minutes = 24,000 seconds.
• The resistance R is thus 24,000 sec / C = 24,000 sec / (.47 Farads) = 24,000 sec / (.47 C / V) = 50,000 ohms.

When t = RC, e^-(t / (R C )) = e^-1 = 1/e, or approximately 1 / 2.7 = .37 of its orginal value.

• The time to fall to half the value is the time t at which e^-(t / (R C)) = 1/2. We can take the natural log of both sides of the equation to obtain t / (R C) = ln(1/2), to that the time t for the voltage default a half its value is R C ln(1/2), or approximately .7 R C.

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The voltage we observe when the magnetic flux through a coil changes is the rate at which the flux changes.

• The magnetic flux through a coil is equal to the product of the component of the magnetic field perpendicular to the coil and the area of the coil.
• The small vector n in red in the figure below is perpendicular to the plane of the coil and is called the normal vector. If `theta is the angle between the magnetic field B and the normal vector, then the component of the magnetic field perpendicular to the loop is B cos(`theta).
• In the figure below, the angle between the magnetic field B and n is `theta = 180 deg. Since cos(180 deg) = -1, we see that the flux is `phi = - B * A.

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In the figure below the loop has been rotated through half a turn so that the normal vector is now parallel and in the same direction to the magnetic field.

• This makes the angle `theta = 0, so that the flux is not equal to B * A.

We see therefore that the flux has changed from - B A to B A, a change of 2 B A.

If the coil was rotated through half a turn in time `dt, then the average rate at which the flux changed is `d`phi / `dt = 2 B A / `dt.

If the coil is rotating with angular frequency `omega, then the angle of the coil with a fixed magnetic field might be `theta = `omega t.

• It would follow that the flux is `phi = B A cos(`theta) = B A cos(`omega t).
• Using calculus we can easily determine the instantaneous rate d`phi / dt at which the flux changes.
• Our result is that V = d`phi / dt = -`omega B A sin(`omega t).
• The graphs of the flux and voltage functions vs. clock time t will therefore be as depicted below, with the phase of the voltage 'leading' that of the flux by 1/4 of a cycle or 90 degrees.

The figure below shows the calculation of the voltage function for this situation. Calculus is used in the third and fourth lines.

• We see that the maximum voltage must be `omega B A.
• From calculus we can easily determine that the average voltage is equal to the maximum voltage divided by `sqrt(2).