Pj Problems - Overview
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An incandescent light bulb rated at 60 W will dissipate 60 W of heat and light when connected across a 100 V ideal voltage source. A 100 W bulb will dissipate 100 W when connected across the same source. If the bulbs are connected in series across the same source, determine the power that either one of the two bulbs will dissipate.
S7P3A32 (force - push).
Pj Problem of Interest is of type force (push).
Equations of interest: V = IR; P = IV; P = I2R
Figure 1.1 is bulb 1 connected to the voltage source.
So, current, I = P/V = 60/100 = 0.6 A.
So, R1 = P/I2 = 60/0.36 = 166.7 Ω
Figure 1.2 is bulb 2 connected to the voltage source.
So, current, I = P/V = 100/100 = 1 A
So, R2 = P/I2 = 100/1 = 100 Ω
Figure 1.3 is both bulbs connected in series to the voltage source.
So, Total circuit resistance, RT = 166.7 + 100 = 266.7 Ω
So, current, I = 100/266.7 = 0.375 A
VR1 = (166.7/266.7)100 = 62.51 V
So, Power dissipated by bulb one = 0.375 x 62.51 = 23.44 W.
VR2 = (100/266.7)100 = 37.5 V
So, Power dissipated by bulb two = 37.5 x 0.375 = 14.06 W.
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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