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 Physical aspects of an electric organ discharge (EOD): Stacks of electrocyte resemble an electrical circuit with in series wired capacitors. The final EOD unit potential is thus the sum of the individual electrocytes: V0 = n×Vn Converting this formula to obtain the number of electrocytes required to achieve the maximal output voltage per EOD-unit yields: n= 50V/0.1V = 500 stacked electrocytes Comparing the discharging pattern of an EOD with that of a capacitor in electronics, some characteristic parameters can be deducted. This type of exponential decrease is very common in nature. The plot of the charge on the capacitor versus time is shown on the right. Assuming that a single peak of an EOD lasts an average of 5ms (if frequency pattern of emitted EODs is 200 Hz), than the time constant (τ) should be significantly less keep remnant electric charge withheld in the electric organ low (would decrease intensity of the peak). Choosing a (τ) of about 1.5ms enables the ray to liberate roughly 99% of the generated voltage into the surrounding environment: VC = V0·(1-e-t/τ) V0, maximal peak voltage generated by the EOD; V0 ≈ 50V t, average duration of a single spike of an EOD pattern; t ≈ 5ms τ, time constant required to recharge an EOD unit; τ = R · C ≈ 1.5ms To improve the volume to capacitance ratio, each EOD unit must have a dielectric (insulating material; usually as fat inclusions or other non conducting tissue deposit). The electric field between the plates of the "capacitor" is weakened by its dielectric. Thus, for a given charge on the plates, the potential difference is reduced and the capacitance (Q/V) is increased considerably. Calculating the capacity of a single EOD-unit reveals a average capacitance of: Ci = κ · ε0 · 4 h · d2 · π ≈ 180[μF] κ, permitivity factor of dielectric other than vacuum [-];κ ≈ 2 (estimated) ε0, dielectric constant 8.8542·E-12 [F/m] h, average height of an EOD-unit (adult T.marmorata); h ≈ 50mm d, diameter of an EOD-unit (adult T.marmorata 0.5m long); d ≈ 5mm Since stacks of electrocyte form one EOD-unit, the capacitance of a single electrocyte can be found as follows:
 Cn = n × Ci  =  72[mF] Ci, Capacitance of an EOD-unit; Ci ≈ 18μF n, number of stacked electrocytes; n ≈ 400 total capacitance of cap's in series: 1/Ci = ∑n/Cn As outlined by the exponential equation above, the decrease is characterized by the time constant, which can be further split to obtain the electrical characteristics, to obtain the internal resistance of each EOD unit, at the moment when the animal emits a series of electrical shocks into the environment: Ri = τ/Ci ≈ 8.3[Ω] τ, time constant of an EOD unit; τ ≈ 1.5ms Ci, capacitance of an EOD unit; Ci ≈ 180μF Keeping in mind that the surrounding tissue of the electric organ does have a specific conductance, implies that the generated peak voltage within the organ must be even greater than the measured one between the ventral and dorsal surfaces of the organ.
 RE = ρ · L ·4d2 · π ≈ 10[Ω] ρ, specific resistivity of epidermis; ρ ≈ 39.3·E-3Ω·m L, average thickness of epidermis; L ≈ 5mm (averaged) d, diameter of an EOD-unit (adult T.marmorata 0.5m long); d ≈ 5mm The resulting peak current of an EOD can be calculated to be roughly (specific resistivity of seawater is neglecible*): Ii = _V0_Ri+RE ≈ 2.7[A] V0, maximal peak voltage generated by the EOD; V0 = 50V Ri, EOD's internal resistance; Ri ≈ 8.3Ω RE, epidermal resistance; RE ≈ 10Ω
 Assuming each lobe of an adult pectoral fin houses roughly 300 individual units, yields a total capacitance of: CT = n · Ci ≈ 108[mF] n, total number of EOD units in an adult T.marmorata; n ≈ 600 Ci, capacitance of an EOD unit; Ci ≈ 180μF Converted to power, the energetic balance of a single EOD can peak to as much as: Pp = V0 · Ii ·n ≈ 81[kW] V0, maximal voltage generated by the EOD; V0 ≈ 50V Ii, peak current of a EOD unit; Ii ≈ 2.7A n, number of EOD units (adult T.marmorata); n ≈ 600 One should keep in mind that this enormous number represents he peaking value for a fraction of a second; thus, power dissipation within a firing interval between peaks is averaged by: Pa = n · (V0·Ii) · et/τ ≈ 2.9[kW] V0, maximal voltage generated by the EOD; V0 ≈ 50V Ii, peak current of a EOD unit ; Ii ≈ 2.7A n, total number of EOD units in an adult T.marmorata; n ≈ 600 t, average peak intervall at a firing rate of 200Hz; t ≈ 5ms τ, time constant required to recharge an EOD unit; τ = R·C ≈ 1.5ms
 Finally, the overall energy stored within both lobes of an adult T.marmorata can be calculated. To make the unit joule more comprehensible, one has to keep in mind that 1[J] is done when a force of 1[N] (0.1kg) is exerted to over a distance of 1m. U = 1/2·CT·V02 ≈ 135 [MJ] CT, total capacitance [C/V] = [F]; CT ≈ 108mF V, voltage [J/C] = [V]; Vmax ≈ 50V The total energy contained in an electric ray such as T.marmorata, is comparable to a capacitor with a capacitance of 110mF at a maximum charging voltage of 25V.  (*) Electrical conductivity of seawater depends upon several parameters such as salinity, temperature, and pressure. For this particular calculation the specific conductivity at T=18°C and 38%0 salinity equals 48·E3/(Ω·m).