alexayero
contestada

A 9-volt battery is used to power two lightbulbs on separate branches of a parallel circuit. The
first light bulb has a resistance of 3 ohms and the second has a resistance of 2 ohms.

What is the power output of the battery?

Please show work if possible!

Respuesta :

Find net resistance first

[tex]\\ \rm\Rrightarrow R=\dfrac{R_1R_2}{R_1+R_2}[/tex]

[tex]\\ \rm\Rrightarrow R=\dfrac{3(2)}{3+2}[/tex]

[tex]\\ \rm\Rrightarrow R=\dfrac{6}{5}[/tex]

[tex]\\ \rm\Rrightarrow R=1.2\Omega[/tex]

Now

[tex]\\ \rm\Rrightarrow P=\dfrac{V^2}{R}[/tex]

[tex]\\ \rm\Rrightarrow P=\dfrac{9^2}{1.2}[/tex]

[tex]\\ \rm\Rrightarrow P=\dfrac{81}{1.2}[/tex]

[tex]\\ \rm\Rrightarrow P=67.5W[/tex]

igoroleshko156

Answer:

[tex]power=67.5\ watts[/tex]

Explanation:

Step 1:  Determine the needed formulas

Net Resistance → [tex]R=\frac{R_1*R_2}{R_1+R_2}[/tex]

Power Output → [tex]P=\frac{V^2}{R}[/tex]

Step 2:  Determine the Net Resistance

[tex]resistance =\frac{3*2}{3+2}[/tex]

[tex]resistance =\frac{6}{5}[/tex]

[tex]resistance =1.2\ \Omega[/tex]

Step 3:  Determine the Power Output

[tex]power=\frac{(9)^2}{1.2}[/tex]

[tex]power=\frac{81}{1.2}[/tex]

[tex]power=67.5\ watts[/tex]

Answer:  [tex]power=67.5\ watts[/tex]