Korzystając ze schematu obwodu, oblicz napięcie na oporze R1.   http://www.kopernik.kalisz.pl/fizyka/img/46.bmp

[latex]U=200 [V] \ R_{3}=30 [ Omega ] \ R_{1}=60 [ Omega ] \ R_{2}=30 [ Omega ] \ I_{1}= frac{U}{R_{z}} = frac{U}{R_{3}+ ( frac{1}{R_{1}} + frac{1}{R_{2}})^{-1}} \ [/latex]   [latex]I_{1}= frac{200 [V]}{30 [ Omega] + ( frac{1}{60 [ Omega ]} + frac{1}{30 [ Omega] }) ^{-1}} = frac{200 [V]}{30 [ Omega ] + 20 [ Omega ]}=4 [A] \ [/latex]   [latex]egin{cases} I_{1}=4 [A] \ I_{1}=I_{2}+I_{3} \ I_{2}R_{1}=I_{3}R_{2} end{cases} \ left { {{4=I_{2}+I_{3}} o I_{3}=4- I_{2} atop {I_{2}R_{1}=I_{3}R_{2}}} ight \ [/latex]   [latex]I_{2}R_{1}=(4- I_{2}) R_{2} \ I_{2}R_{1}=4R_{2}-I_{2}R_{2} \ I_{2} (R_{1}+R_{2})=4R_{2} \ I_{2}= frac{4R_{2}}{R_{1}+R_{2}}= frac{4 cdot 30 [ Omega ]}{90 [ Omega ]}=1 frac{1}{3} [A] \ [/latex]   [latex]U_{R_{1}}=I_{2}R_{1}=1 frac{1}{3} [A] cdot 60 [ Omega ] =80 [V][/latex]