Unformatted text preview: The bulb in arrangement (c) is the brightest because the batteries are wired in series and the voltage is twice the voltage of a single battery. Exercises 2. We solve this exercise using Ohm’s law. 120 V 8 15 A V R I = = = Ω If the resistance is lower than 8 Ω , the current will be even larger than 15 A. 8. We first need to realize that connecting two batteries in parallel does not increase the voltage. Therefore, the voltage is 1.5 V. We now use Ohm’s law to find the current through the resistor. 1.5 V 0.5 A 3 V V R I I R = ⇒ = = = Ω Because the batteries are connected in parallel, half the current flows through each battery. Therefore, the current through each battery is 0.25 A....
View
Full Document
 Fall '02
 GregoryFrancis
 Batteries, Volt, Resistor, Electrical resistance

Click to edit the document details