Question

ELECTRICITATE..........

Answer 1

 

$\displaystyle\bf\\Se~da:\\R_G=4,\!9\Omega\\N=50~diviziuni\\I_G=100mA\\Se~cere:\\R_{sunt}\\In~conditiile:\\I_1=2A~deviaza~acul~cu~N_1=20~de~diviziuni.\\\\Rezolvare:\\\\\text{Calculam curentul care deviaza acul la maxim, adica 50 de diviziuni}\\2A................ 20~de diviziuni\\I.................. 50~de diviziuni\\\\I=\frac{2\times50}{20}=\frac{100}{20}=5A\\\\I=5A$

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$\displaystyle\bf\\\text{Calculam raportul curentilor:}\\\\n=\frac{I}{I_G}=\frac{5A}{100mA}=\frac{5A}{0,1A}=50\\\\\text{Calculam suntul:}\\\\R_{sunt}=\frac{R_G}{n-1}=\frac{4,\!9}{50-1}=\frac{4,\!9}{49}=0,\!1\Omega\\\\\boxed{\bf~R_{sunt}=0,\!1\Omega}\\\\\boxed{\bf~Raspuns~corect:~b)}$