In this circuit configuration, if the resistor R1 is in series with LED1, then the current flowing through R1 will be the same as the current flowing through LED1.
Since you are given that the current through R1 is 5 mA, the current flowing through LED1 would also be:
\[ \text{Current through LED1} = \text{Current through R1} = 5 , \text{mA} \]
Therefore, the current flowing through LED1 is 5 mA.
1 answer