Vraag 3

uitkomst:R=1{,}7\cdot10^2\,\OmegaR=1{,}7\cdot10^2\OmegaR=1{,}7\cdot10^2\Omega$R=1{,}7 \cdot 10^{2} \Omega(met een marge van0{,}1\cdot10^2\,\Omega0{,}1\cdot10^2\Omega0{,}1\cdot10^2\Omega$0{,}1 \cdot 10^{2} \Omega)

voorbeeld van een berekening: Voor de serieschakeling geldt:$U_{\text {batt }}=U_{\mathrm{R}}+U_{\text {rood }}+U_{\text {groen }}+U_{\text {blauw }}=9{,}0 \mathrm{~V}.

Bij10\text{ mA}101010101010101010101010mlezen we de spanning over de LEDs af:

U_{\text{rood }}=1{,}85~\text{V};\,U_{\text{groen }}=2{,}57~\text{V};\,U_{\text{blauw }}=2{,}85~\text{V}U_{\text{rood }}=1{,}85~V\text{V};\,U_{\text{groen }}=2{,}57~\text{V};\,U_{\text{blauw }}=2{,}85~\text{V}U_{\text{rood }}=1{,}85~V;\,U_{\text{groen }}=2{,}57~\text{V};\,U_{\text{blauw }}=2{,}85~\text{V}U_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~\text{V};\,U_{\text{blauw }}=2{,}85~\text{V}U_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~\text{V};\,U_{\text{blauw }}=2{,}85~\text{V}U_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~\text{V};\,U_{\text{blauw }}=2{,}85~\text{V}U_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~\text{V};\,U_{\text{blauw }}=2{,}85~V\text{V}U_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~\text{V};\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V\text{V};\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;\,U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;U_{\text{blauw }}=2{,}85~VU_{\text{rood }}=1{,}85~V;U_{\text{groen }}=2{,}57~V;U_{\text{blauw }}=2{,}85~V$U_{\text {rood }}=1{,}85 \mathrm{~V} ; U_{\text {groen }}=2{,}57 \mathrm{~V} ; U_{\text {blauw }}=2{,}85 \mathrm{~V}.

Hieruit volgt:U_{R}=9{,}0-(1{,}85+2{,}57+2{,}85)=1{,}73~\text{V}U_{R}=9{,}0-(1{,}85+2{,}57+2{,}85)=1{,}73~$U_{\mathrm{R}}=9{,}0-(1{,}85+2{,}57+2{,}85)=1{,}73 \mathrm{~V}.

Voor de stroom door R geldt:I=0{,}010~\text{A}I=0{,}010~\text{VA}I=0{,}010~\text{V}I=0{,}010~$I=0{,}010 \mathrm{~A}.

Hieruit volgt:R=\frac{U}{I}=\frac{1{,}73}{0{,}010}=173\,\Omega=1{,}7\cdot10^2\,\OmegaR=\frac{U}{I}=\frac{1{,}73}{0{,}010}=173\,\Omega=1{,}7\cdot10^2\OmegaR=\frac{U}{I}=\frac{1{,}73}{0{,}010}=173\,\Omega=1{,}7\cdot10^2\OmegaR=\frac{U}{I}=\frac{1{,}73}{0{,}010}=173\,\Omega=1{,}7\cdot10^2\OmegaR=\frac{U}{I}=\frac{1{,}73}{0{,}010}=173\Omega=1{,}7\cdot10^2\OmegaR=\frac{U}{I}=\frac{1{,}73}{0{,}010}=173\Omega=1{,}7\cdot10^2\Omega$R=\frac{U}{I}=\frac{1{,}73}{0{,}010}=173 \Omega=1{,}7 \cdot 10^{2} \Omega. .