# The Blame

Topics: Volt, Electric current, Battery Pages: 1 (267 words) Published: March 9, 2013
Vlll. Questions
1. A cell of emf 3.0 volts and internal resistance of 0.01 ohm is connected through an ammeter of 0.05-ohm internal resistance to a 5.0Ω rheostat by wire having a total resistance of 0.85-ohm. In the calculation of the current, what percentage error would be made by neglecting all resistance except that of the rheostat?

2. A dry cell when short-circuited will furnish about 30 amperes for a brief time. If its emf is 1.5 volts, what is its internal resistance? An ordinary household electric lamp takes about 1 amp will it be safe to connect it directly to a dry cell? Why? Answer: The internal resistance = 1.5/30=0.05 ohm.

The household electric works on much higher voltage, so enough number of dry cells much be put in service to meet the voltage demand. For example the rated voltage of the lamp is 110 Volts 73 dry cells must be connected in series to form a battery. The internal resistance of the battery thus formed would be 0.05 x 73= only 3.65 ohms.So it is possible to light the bulb. R = E/I, = 110/1, = 110 ohms as a hot resistance for the filament. Connecting it to a single 1.5V cell would be perfectly safe, even if the cold resistance of the lamp is considerably less than its hot resistance. To get 1A to flow and the lamp to light, the cell would need to have a voltage of 110V., or you would need 74 cells in series.