An ordinary flashlight uses two D cells 1.50 V batteries connected in series. The bulb draws 380 m when turned on.

a. Calculate the resistance of the bulb and the power dissipated.
b. By what factor would the power increase if four D cells in series were used with the same bulb? (Neglect heating effects of the filament.) Why shouldn’t you try this?

a. 7.89 Ω, 1.14 W

b. By a factor of 2; The batteries would heat up.

Explanation:

a. Using Ohm's law, we have that Voltage, V, is directly proportional to Current, I, with the constant of proportionality being Resistance, R. Mathematically:

V = IR

The voltage of the two batteries would be:

V = V1 + V2 = 1.5 + 1.5 = 3.0 V

The current is 380 mA = 0.38 A

Hence, the Resistance, R, will be:

R = 7.89 Ω

Power is given as:

P = IV

P = 3 * 0.38

P = 1.14 W

b. If the batteries are now 4, the new voltage will be:

V = 4 * 1.5

V = 6 V

Power becomes:

P = 0.38 * 6

P = 2.28 W

Comparing with Power in (a) above, we see that the new Power is double the value of the former Power. Hence, the Power has increased by a factor of 2.

It is not advisable to use double the number of batteries to power a flashlight because it would cause the batteries to heat up and thereby, leak or in the worst case scenario, blow up. This could be damaging to the flashlight.

im sorry i dont know

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2