Answer:
The power is [tex]P_p = 4.6 \ W[/tex]
Explanation:
From the question we are told that
The power delivered is [tex]P_{s} = 1.15 \ W[/tex]
Let it resistance be denoted as R
The resistors are connected in series so the equivalent resistance is
[tex]R_{eqv} = R+ R = 2 R[/tex]
Considering when it is connected in series
Generally power is mathematically represented as
[tex]P_s = V * I[/tex]
Here I is the current which is mathematically represented as
[tex]I = \frac{V}{2R}[/tex]
The power becomes
[tex]P_s = V * \frac{V}{2R}[/tex]
[tex]P_s = \frac{V^2}{2R}[/tex]
substituting value
[tex]1.15 = \frac{V^2}{2R}[/tex]
Considering when resistance is connected in parallel
The equivalent resistance becomes
[tex]R_{eqv} = \frac{R}{2}[/tex]
So The current becomes
[tex]I = \frac{V}{\frac{R}{2} } = \frac{2V}{R}[/tex]
And the power becomes
[tex]P_p = V * \frac{2V}{R} = \frac{2V^2}{R} = \frac{4 V^2}{2 R} = 4 * P_s[/tex]
substituting values
[tex]P_p = 4 * 1.15[/tex]
[tex]P_p = 4.6 \ W[/tex]
