Korzystamy ze wzoru na opór elektryczny:
[tex]R = \frac{U}{I}[/tex]
[tex]a)\\U = 2 \ V\\I = 0,5 \ A\\\\R = \frac{2 \ V}{0,5 \ A} = 4 \ \Omega[/tex]
[tex]b)\\U = 5 \ V\\I = 20 \ mA = 0,02 \ A\\\\R = \frac{5 \ V}{0,02 \ A}= 250 \ \Omega[/tex]
[tex]c)\\U = 100 \ mV = 0,1 \ V\\I = 2 \ mA = 0,002 \ A\\\\R = \frac{0,1 \ V}{0,002 \ A} = 50 \ \Omega[/tex]
[tex]d)\\U = 5 \ V\\I = 1 \ A\\\\R = \frac{5 \ V}{1 \ A} = 5 \ \Omega}[/tex]
