This relates to the following two questions: I he circuit in the figure consists

Question

This relates to the following two questions: I he circuit in the figure consists of a battery, a thick-filament bulb (R|). a thin-filament bulb , approximately how many electrons How out the battery every second at location A in the steady state?

Explanation / Answer

R = resistance = resistivity * length / Cross sectional Area Resistivity and area are given to be same for both filaments. The thicker filament would have greater cross sectional area and thus lower resistance. So, R2>R1. In a series circuit current through both filaments will be same. So, the one with higher resistance will experience greater potential drop and power dissipation and thus will glow brighter. So, the answer is R2. If the diameter of R1 is twice that of R2, R1~R2/4 since area is proportional to square of diameter. So, in a series circuit, Rnet = R1+R2 = 5R1 ~ 10 ohm. So, Current = I = V/Rnet = 16/10 A = 1.6 A. Since electronic charge = 1.6*10^-19 C, no. of electrons passing per second = 10^19.

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