Ohm's law and electric currents problem? Please help!?
A 47Ω resistor can dissipate up to 0.22 W of power without burning up. What is the smallest number of such resistors that can be connected in series across a 9.0 V battery without any one of them burning up?
- billrussell42Lv 79 years agoFavorite Answer
P = E²/R
for one resistor
0.22 = E²/47
E = 3.22 volts
9v / 3.22v = 2.80
so the answer is 3.
they will have 3 volts across them
P = 9/47 = 0.191 watts OK.
- rigelLv 44 years ago
R = E / I, sparkling up for R A = rho * L / R you know the resistivity of copper (a million.sixty 8×10?8), you know the dimensions, and by fixing the formula above, you are going to know what the area must be. area = pi * r^2, sparkling up for r