LED resistor current

Calculate the current I that would race through resistor R5 (720 ohms), resistor R2 (220 ohms), and an LED (U=2 V) at a DC 6 V source. The resistors are connected in series.

Final Answer:

I =  4.2553 mA

Step-by-step explanation:

$$\begin{gathered} R_5=720\Omega \\ {R}_2=220\Omega \\ {U}_3=2\text{V} \\ U=6\text{V} \\ {U}_2=U-U_3=6-2=4\text{V} \\ R=R_2+R_5=220+720=940\Omega \\ {I}_2=\frac{U_2}{R}=\frac{4}{940}=\frac{1}{235}\doteq 0.0043\text{A} \\ I=I_2\cdot 1000=0.0043\cdot 1000=4.2553\text{mA} \end{gathered}$$

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