Answer
The ratio of the total power delivered when the resistors are connected in parallel to the total power delivered when the resistors are connected in series is $~9.5$
Work Step by Step
We can find the equivalent resistance when the resistors are connected in parallel:
$\frac{1}{R_{eq}} = \frac{1}{2.5~k\Omega} + \frac{1}{3.5~k\Omega} + \frac{1}{4.5~k\Omega}$
$\frac{1}{R_{eq}} = \frac{126}{315~k\Omega} + \frac{90}{315~k\Omega} + \frac{70}{315~k\Omega}$
$\frac{1}{R_{eq}} = \frac{286}{315~k\Omega}$
$R_{eq} = \frac{315~k\Omega}{286}$
$R_{eq} = 1101.4~\Omega$
We can find the total power delivered:
$P = \frac{V^2}{R_{eq}} = \frac{(100~V)^2}{1101.4~\Omega} = 9.0794~W$
We can find the equivalent resistance when the resistors are connected in series:
$R_{eq} = 2.5~k\Omega + 3.5~k\Omega + 4.5~k\Omega$
$R_{eq} = 10.5~k\Omega$
We can find the total power delivered:
$P = \frac{V^2}{R_{eq}} = \frac{(100~V)^2}{10,500~\Omega} = 0.9524~W$
We can find the ratio of the total power delivered:
$\frac{9.0794~W}{0.9524~W} = 9.5$
The ratio of the total power delivered when the resistors are connected in parallel to the total power delivered when the resistors are connected in series is $~9.5$
