A circuit with 5v connected to a 2kΩ resistor. Series to the 2kΩ resistor is connected a Silicondiode and a 1kΩ resistor in parallel. What is current through 2kΩ resistor?
Question
A circuit with 5v connected to a 2kΩ resistor. Series to the 2kΩ resistor is connected a Silicondiode and a 1kΩ resistor in parallel. What is current through 2kΩ resistor?
Solution
To find the current through the 2kΩ resistor, we need to consider the entire circuit.
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First, we need to understand that a silicon diode has a voltage drop of approximately 0.7V when forward biased.
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The total voltage in the circuit is 5V. When the silicon diode is forward biased, it will drop 0.7V, leaving 5V - 0.7V = 4.3V across the resistors.
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The resistors are in series, so the total resistance in the circuit is 2kΩ + 1kΩ = 3kΩ.
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Ohm's law states that I = V/R. So, the current through the 2kΩ resistor is I = 4.3V / 3kΩ = 1.43 mA.
So, the current through the 2kΩ resistor is approximately 1.43 mA.
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