Câu 4:
Ta có:
\(\begin{array}{l}
R = \dfrac{{2x}}{{x + 2}} + \left( {6 - x} \right) = \dfrac{{ - {x^2} + 12}}{{x + 2}}\\
I = \dfrac{U}{R} = \dfrac{{10\left( {x + 2} \right)}}{{ - {x^2} + 12}}\\
{I_A} = \dfrac{x}{{x + 2}}.\dfrac{{10\left( {x + 2} \right)}}{{ - {x^2} + 12}} = \dfrac{{10x}}{{ - {x^2} + 12}} = 1\\
\Rightarrow x = 1,08\Omega
\end{array}\)
Số chỉ vôn kế là:
\({U_V} = U - {I_A}{R_1} = 10 - 1.2 = 8V\)
Câu 5:
a) Ta có: \(\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} \Rightarrow \left\{ \begin{array}{l}
R < {R_1}\\
R < {R_2}
\end{array} \right.\)
b) Ta có:
\(\begin{array}{l}
\dfrac{1}{{{R_{td}}}} = \dfrac{1}{R} + \dfrac{1}{R} + ... + \dfrac{1}{R} = \dfrac{n}{R}\\
\Rightarrow {R_{td}} = \dfrac{R}{n} \Rightarrow 2 = \dfrac{{50}}{n} \Rightarrow n = 25
\end{array}\)
