91影视

Consider the following circuit, where${蔚}_1$=20.0 V, ${蔚}_2$=200 V, R鈧�=8.00 $惟$, R鈧�= 18.0 $惟$, $R_3$= 30.0$惟$and $R_4$=20.0 $惟$. The current in the 20.0 V battery is I= 5.00 A in the direction shown and the voltage across the R鈧� resistor is V R鈧� = 16.0 V, with the lower end of the resistor at higher potential.

(a) First we need to find the emf $蔚$, to do that we simplify the resistors networks, as shown in the second figure, we know that the resistance add in series, and the reciprocal of the resistance add in parallel, so we have,

$$\begin{gathered} \frac{1}{R^{\mathrm{鈥�}}}=\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{30.0\mathrm{惟}}+\frac{1}{20.0\mathrm{惟}}=\frac{1}{12.0\mathrm{惟}} \\ {R}^{\mathrm{鈥�}}=12.0\mathrm{惟} \end{gathered}$$ (1)

and the resistance of the two resistors each with the resistance of R is R/2 Now we apply the loop rule to the left loop (counterclockwise), so we can write (note that we have assumed that the lower end of the battery is the negative pole, if we get a positive value then our assumption will be correct if negative we must reverse its polarity),

$蔚-V{R}_1-I(R\text{'}+R_2)-{蔚}_1=0$

or,

$蔚=V{R}_1+I(R\text{'}+R_2)+{蔚}_1$

substitute to get

$蔚=16.0V+(5.00A)(12.0惟+18.0惟)+20.0V=186V$

$蔚=186V$

with the upper terminal being the positive

(b) From the junction rule, we can write,

$I_2=I-I_1$

we know the current I, the current I₁ can be determined from the resistor R₁ that is $I_1=V_{R}1/R1$, thus,

$I_2=I-\frac{V_{R1}}{R1}$

substitute to get,

localid="1668337436073" $$\begin{gathered} I_2=5.00A-\frac{16.0V}{8.00惟}=3.00A \\ {I}_2=3.00A \end{gathered}$$'

(c)Now we need to find the resistance R and apply the loop rule to the right loop (counterclockwise), so we get,

$$\begin{gathered} {蔚}_2鈭�\frac{I_{2}R}{2}+V_{R_1}鈭�蔚=0 \\ R=\frac{2{蔚}_2+2V_{R_1}鈭�2蔚}{I_2} \end{gathered}$$

substitute to get,

$$\begin{gathered} R=\frac{2\left(200V\right)+2\left(16.0\mathrm{V}\right)鈭�2\left(186\mathrm{V}\right)}{3.00\mathrm{A}}=20.0\mathrm{惟} \\ \mathrm{R}=20.0\mathrm{惟} \end{gathered}$$