Consider the two circuits shown in the figure below in which the batteries are i

Question

Consider the two circuits shown in the figure below in which the batteries are identical. The resistance of each lightbulb is R. Neglect the internal resistance of the batteries. (a) Find expressions for the currents in each lightbulb. (Use the following as necessary: for epsilon and R.) I_A = epsilon/R I_B = epsilon/2R I_C = epsilon/2R (b) How does the brightness of B compare with that of C? (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) B=C Explain. since they both have the same current and voltage they will also have the same brightness (c) How does the brightness of A compare with that of B and of C? (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) A > B = C Explain. A will be brighter then B and C because it has more current going through it, and the same amount of voltage, therefore it has greater power and will be brighter

Explanation / Answer

A) According to ohm's law

current i = V/R

v is the emf of the battery

then

IA = e/R

for the second circuit with two bulbs

net resistance is Rnet = R1+R2 = R+R = 2R

IB = e/(2R)

Ic = e/(2R)

your inputs are correct only

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