Answer:
P(bat) = V²r/(R+r)²
Explanation:
Let the resistance of the coil be R
Internal resistance of the battery be r
Emf of the battery = V
Power dissipated in the internal resistance of the battery is normally given as P = I²r
where I is the current flowing in the circuit.
From Ohm's law,
V = I R(eq)
R(eq) = (R + r)
I = V/(R+r)
P = I²r
P = [V/(R+r)]²r
P = V²r/(R+r)²
Hope this Helps!!!
Answer:
The answer is 4x³ + 6x²
<u>-TheUnknownScientist</u><u> 72</u>
V = [4/3]π r^3 => [dV / dr ] = 4π r^2
[dV/dt] = [dV/dr] * [dr/dt]
[dV/dt] = [4π r^2] * [ dr/ dt]
r = 60 mm, [dr / dt] = 4 mm/s
[dV / dt ] = [4π(60mm)^2] * 4mm/s = 180,955.7 mm/s