- #1
johnson.3131
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A light bulb is wired in series with a 132 resistor, and they are connected across a 120 V source. The power delivered to the light bulb is 23.4 W. What are the two possible resistances of the light bulb?
a. lower value
b. higher value
relevant equations V=I/R and P=(I^2)(R)
attempt at soultion:
23.4=I^2(132)
=.4210
Wrong!
V=I/R
120=I/132
=15840
Wrong!
a. lower value
b. higher value
relevant equations V=I/R and P=(I^2)(R)
attempt at soultion:
23.4=I^2(132)
=.4210
Wrong!
V=I/R
120=I/132
=15840
Wrong!