a) The formula for calculating the quantity of charge is expressed as
Q = IT
where
Q is the quantity of charge
I is the current
T is the time
From the information given,
I = 10.3
T = 5 ms = 5 x 10^-3 s
Q = 10.3 x 5 x 10^-3
Q = 51.5 x 10^- 5 C
The quantity of charge passed is 51.5 x 10^- 5 C
b) The formula for calculating the energy is expressed as
E = I^2RT
where
R is the resistance
E is the energy
From the information given,
E = 492 J
Thus,
492 = 10.3^2 x R x 5 x 10^-3
R = 492/(5 x 10^-3 x 10.3^2)
R = 927.514 ohms
Voltage, V = IR
Voltage = 10.3 x 927.514
Voltage = 9553.398 V
We would divide by 1000. It becomes
Voltage = 9.553 KV
c) From the calculations,
Resistance = 927.514 ohms
We would divide by 1000. It becomes
Resistance = 0.93 ΚΩ
d) Let the temperature increase be t
mass of tissue, m = 8 kg
Specific heat of tissue = 3500 J/(kg. °C).
°C
The formula for calculating the quantity of heat is
H = mcθ
where
H is the quantity of heat
From the informtaion given,
H = 492
θ = t
Thus,
492 = 8 x 3500 x t
t = 492/(8 x 3500)
t = 0.018
The temperature increase is 0.018 degrees