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?
Answer:
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:
[tex]R = \frac{V}{I}[/tex]
[tex]R = \frac{3}{0.38}[/tex]
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.