Question

Figure shows three light bulbs connected to a 120-V AC (rms) household supply voltage. Bulbs 1 and 2 have a power rating of 150-W , and bulb 3 has a 100-W rating. Then,

• A. the rms current in 150-W bulbs is 1.25 A
• B. the rms current in 150-W bulbs is 0.833 A
• C. the rms current in 100-W bulb is 0.833 A
• D. total resistance of the combination of the three light bulbs is 36 Ω

Answer 1

Answer: (A), (C), (D)

total resistance of the combination of the three light bulbs is 36 Ω

All the lamps are connected in parallel with the voltage source,
$\Delta V_{rms}=120V$ for each lamp. Also, the current is $I_{rms}=P_{avg}$ / $\Delta V_{rms}$
and the resistance is $R=\Delta V_{rms}/I_{rms}$

For the 150-W bulbs, $I_{rms}=\frac{150W}{120V}=1.25A$

For the 100-W bulb, $I_{rms}=\frac{100W}{120V}=0.833A$
The resistance of bulbs 1 and 2 is
$R_1=R_2=\frac{120V}{1.25A}=96\Omega$

and the resistance of bulb 3 is
$R_3=\frac{120V}{0.833A}=144\Omega$
Now the total resistance all combined is
$\frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
$R_{eff}=36\Omega$