Question

Help me with this one.

[ In attachment below ]

Answer 1

@Festinger

Answer 2

Let each wire have resistance R since they are similar.

When connected in series, total resistance is 2R.
Since power is
\[P=\frac{V^{2}}{R}\]Power dissipated in this circuit is:
\[P_{1}=\frac{V^{2}}{2R}\]

Now, it is connected in parallel.
\[\frac{1}{R_{total}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\]
So total resistance is R/2

\[P_{2}=\frac{2V^{2}}{R}\]

Ratio is:
\[P_{1}:P_{2}\]
\[\frac{V^{2}}{2R}:\frac{2V^{2}}{R}\]
\[\frac{1}{2}:2\]
\[1:4\]

Answer 3

i don't understand

Answer 4

@Fifciol can you explain me?

Answer 5

Let's first take in series:

Power dissipated
\[P_1=\frac{ V^2 }{ R_{eq} }=\frac{ V^2 }{ 2R} \]
we cannot take \[P=I^2R \]
because voltage remains the same therefore current must change
in parallel:

\[\frac{ 1 }{ R_{eq} }=\frac{ 1 }{ R }+\frac{ 1 }{ R }=\frac{ 2 }{ R }\rightarrow R_{eq}=\frac{ R }{ 2 }\]
\[P_2=\frac{ V^2 }{ \frac{ R }{ 2} }=V^2*\frac{ 2 }{ R }=\frac{ 2V^2 }{ R }\]
\[\frac{ P_1 }{P_2 }=\frac{ \frac{ V^2 }{ 2R } }{ \frac{2V^2 }{ R} }=\frac{ V^2 }{ 2R }*\frac{ R }{ 2V^2 }=\frac{ 1 }{ 4 }=1:4\]