91影视

The equivalent resistance for a series and parallel combination of resistors can be expressed as,

$$\begin{gathered} {\mathit{R}}_{\mathbf{s}\mathbf{e}\mathbf{r}\mathbf{i}\mathbf{e}\mathbf{s}}\mathbf{=}\underset{\mathbf{i}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{\mathbf{鈭�}}}{\mathit{R}}_{\mathbf{i}} \\ \frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{a}\mathbf{l}\mathbf{l}\mathbf{e}\mathbf{l}}}\mathbf{=}\underset{\mathbf{j}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{\mathbf{鈭�}}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{j}}} \end{gathered}$$

Consider three identical resistors which are connected in series. Consider a certain potential difference that is applied across the combination, the total power dissipated is 450 W. Let R be the resistance of each one of the resistors, then the total power dissipated in the resistors is

$P=\frac{V^2}{R_{eq}}$ (1)

Where$R_{eq}^{series}=3R$so we get from equation (1), that

$$\begin{gathered} P_{series}=\frac{V^2}{R_{eq}^{series}} \\ {P}_{series}=\frac{V^2}{3R} \\ \frac{V^2}{R}=3P_{series} \end{gathered}$$ (2)

Where the value of$P_{series}$ is45.0 W.

Now we need to find out the total power dissipated when the resistors are connected in parallel, the equivalent resistance of the resistors when they are connected in parallel that

$R_{eq}^{paralle}=\frac{R}{3}$

Hence the power dissipated in these resistors is,

$$\begin{gathered} P_{parallel}=\frac{V^2}{R_{eq}^{parallel}} \\ {P}_{parallel}=\frac{3V^2}{R} \end{gathered}$$

using equation (2)

$$\begin{gathered} P_{parallel}=9P_{series} \\ {P}_{parallel}=9脳45W \\ {P}_{parallel}=405W \end{gathered}$$

Hence the power dissipated is 405 W when they are connected in parallel.