91影视

Consider the following circuit (the left circuit), where R = 5.00 惟, all the resistors are rated at a maximum power of 2.00 W.

The current that flows through the resistors 5R, 6R, and 8R are the same (say I) since they are connected in series, and for the resistors connected in parallel, the current is less than this current.

Since the power of the resistor is P = I虏R, then the resistor with the maximum resistance of the series resistors (5R, 6R, and 8R, note that the equivalent resistance of the parallel resistors is less than 8R) reaches the maximum power before the others (since the power depends on the resistance also), therefore the resistor 8R reaches the maximum power of 2.00 W, and after that, no increase in the power is allowed.

We need to find the current associated with this power,

$$\begin{gathered} I=\sqrt{\frac{P}{8P}} \\ =\sqrt{\frac{2.00W}{8\left(5.00惟\right)}} \\ =0.224A \end{gathered}$$ (1)

this is the current that flows in the battery, to find the emf of the battery, we need to find the equivalent resistance first, we know that the resistance add in series and its reciprocal is added in parallel, so we can write (consider the second figure where the resistance of the two parallel resistors of the same resistance is half of one of them),

$$\begin{gathered} \frac{1}{R_{parallel}}=\frac{1}{5R}+\frac{1}{3R}+\frac{1}{4R} \\ \frac{1}{R_{parallel}}=\frac{47}{60R} \\ {R}_{parallel}=\frac{60}{47}R \end{gathered}$$

thus,

$$\begin{gathered} R_{equivalent}=\frac{60R}{47}+5R+5R+6R+8R \\ =\frac{1188R}{47} \end{gathered}$$

the emf is, therefore,

$$\begin{gathered} 蔚=IR \\ =1脳\frac{1188R}{47} \\ =\frac{1188\left(0.224A\right)\left(5.00惟\right)}{47} \\ =28.3V \end{gathered}$$