You should now be comfortable with charge, current, voltage, power, and energy. A 12kA lightning bolt strikes an object directly for 15 µs. Consider the following: a. How much charge is deposited on the object? b. How many electrons is this? (For comparison, there are approximately 1028 electrons in the human body.) c. If the lightning strike discharges across a voltage potential of 250 million volts, what is the lightning bolt’s power? (Is this more or less than 1.21 gigawatts?) d. How much energy is imparted by the lightning strike? e. A typical cell phone battery has a charged capacity of 1800 mAh at 3.7 V. If the energy from the lightning strike were harnessed,

Correct Question: You should now be comfortable with charge, current, voltage, power, and energy. A 12kA lightning bolt strikes an object directly for 15 µs. Consider the following: a. How much charge is deposited on the object? b. How many electrons is this? (For comparison, there are approximately 1028 electrons in the human body.) c. If the lightning strike discharges across a voltage potential of 250 million volts, what is the lightning bolt’s power? (Is this more or less than 1.21 gigawatts?) d. How much energy is imparted by the lightning strike? e. A typical cell phone battery has a charged capacity of 1800 mAh at 3.7 V. If the energy from the lightning strike were harnessed, how many cell phone batteries could be fully charged using this energy?

(a) 0.18 C

(b) 1.125 × 10¹⁸ electrons

(c) 3000 gigawatts ( more than 1.21 gigawatts)

(d) 4.5×10⁷ J

(e) 1877 batteries

Explanation:

(a)

Q = it Equation 1

Where Q = charge deposited on the object, i = current of the lightning bolt, t = duration of strike.

Given: i = 12 kA = 12×10³ A, t = 15 µs = 15×10⁻⁶ s.

Substitute into equation 1

Q = 12×10³(15×10⁻⁶)

Q = 0.18 C

(b)

If the charge of one electron is approximately 1.602 × 10⁻¹⁹ C,

Then,

Number of electrons = 0.18/(1.602 × 10⁻¹⁹ )

Number of electrons = 1.125 × 10¹⁸ electrons

(c)

Power of the lightning bolt(P) = Voltage(V)×current(I)

P = VI Equation 2

Given: V = 250 million volts = 250×10⁶ V, I = 12 kA = 12×10³ A

Substitute into equation 2

P = 250×10⁶(12×10³)

P = 3×10¹² W.

P = 3000 gigawatts ( more than 1.21 gigawatts)

(d)

Energy impacted(E) = Power(P)×Time(t)

E = Pt equation 3

Given: P = 3×10¹² W, t = 15 µs = 15×10⁻⁶ s

Substitute into equation 3

E = 3×10¹²(15×10⁻⁶)

E = 4.5×10⁷ J

(e)

Number of cell phone battery that could be charged(n') = Energy of the lightning bolt(E)/Energy one cell phone battery (E')

n' = E/E' Equation 4

Given: E =  4.5×10⁷ J, E' = 1800×3.7×3600/1000 = 23976 J

Substitute into equation 4

n' =  4.5×10⁷/23976

n' = 1876.88  batteries

n' ≈ 1877 batteries

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