Question

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!

Answer 1

Find net resistance first

`\\ \rm\Rrightarrow R=(R_1R_2)/(R_1+R_2)`

`\\ \rm\Rrightarrow R=(3(2))/(3+2)`

`\\ \rm\Rrightarrow R=(6)/(5)`

`\\ \rm\Rrightarrow R=1.2\Omega`

Now

`\\ \rm\Rrightarrow P=(V^2)/(R)`

`\\ \rm\Rrightarrow P=(9^2)/(1.2)`

`\\ \rm\Rrightarrow P=(81)/(1.2)`

`\\ \rm\Rrightarrow P=67.5W`

Answer 2

`power=67.5\ watts`

Step-by-step explanation:

Step 1: Determine the needed formulas

Net Resistance →
`R=(R_1*R_2)/(R_1+R_2)`

Power Output →
`P=(V^2)/(R)`

Step 2: Determine the Net Resistance

`resistance =(3*2)/(3+2)`

`resistance =(6)/(5)`

`resistance =1.2\ \Omega`

Step 3: Determine the Power Output

`power=((9)^2)/(1.2)`

`power=(81)/(1.2)`

`power=67.5\ watts`

Answer:
`power=67.5\ watts`