To calculate the current flowing through the resistor \( R1 \) in the circuit described, we can use Ohm's Law, which states that:
\[ I = \frac{V}{R} \]
Where:
- \( I \) is the current (in Amperes),
- \( V \) is the voltage across the resistor (in Volts), and
- \( R \) is the resistance (in Ohms).
In the given circuit:
- The supply voltage \( V1 \) is 5V.
- The resistance \( R1 \) is 1kΩ (which is 1000Ω).
Assuming that the LED is connected in series with resistor \( R1 \) and that it does not significantly affect the voltage across \( R1 \) for the sake of this calculation (either by being in a forward conducting state or considered negligible for voltage drop), we can take the entire supply voltage for the drop across \( R1 \).
Calculation
- Substitute the known values into Ohm's Law:
\[ I = \frac{V}{R} = \frac{5V}{1000Ω} \]
- Perform the calculation:
\[ I = \frac{5}{1000} = 0.005A = 5mA \]
Result
The value of the current flowing through \( R1 \) is 5mA.
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