The maximum rated power of the LED is $2 \text{ mW}$ and it is used in the circuit with input voltage of $5 \text{ V}$ as shown in the figure below. The current through resistance $R$ is $0.5 \text{ mA}$. The minimum value of the resistance $R_s$, to ensure that the LED is not damaged is ______ $\text{k}\Omega$.
In the provided circuit, the input voltage is $5 \text{ V}$, a series resistor $R_s$ is connected, and the circuit then branches into two parallel paths: one with a resistor $R = 1 \text{ k}\Omega$ in series with a Zener diode, and another with the LED.
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The total current is the sum of the LED current and the current through $R$. Use $P = VI$ for the LED and assume a standard operating voltage of $2\text{V}$ to find the current.
To prevent damage to the LED, the power dissipated by it should not exceed its maximum rating of $2 \text{ mW}$. The LED and the parallel branch containing resistor $R$ are connected across the same node voltage, let's call it $V_{node}$.
First, identify the voltage across the parallel combination. In such standard problems, a red LED typically has a forward operating voltage $V_{LED}$ of approximately $2 \text{ V}$. Let's verify this against the given options. If $V_{LED} = 2 \text{ V}$, then the power constraint is: $$P_{LED} = V_{LED} \times I_{LED} \le 2 \text{ mW}$$ For the maximum safe limit, we set $P_{LED} = 2 \text{ mW}$: $$I_{LED} = \frac{2 \text{ mW}}{2 \text{ V}} = 1 \text{ mA}$$ We are given that the current through resistor $R$ is $I_R = 0.5 \text{ mA}$. Since this branch is in parallel with the LED, the total current $I_{total}$ flowing through the series resistor $R_s$ is the sum of the currents in the two branches: $$I_{total} = I_{LED} + I_R = 1 \text{ mA} + 0.5 \text{ mA} = 1.5 \text{ mA}$$ Now, apply Kirchhoff's Voltage Law (KVL) to the outer loop of the circuit. The sum of the voltage drops across $R_s$ and the parallel branch must equal the source voltage $V_{in} = 5 \text{ V}$: $$V_{in} = I_{total} R_s + V_{LED}$$ $$5 \text{ V} = (1.5 \text{ mA}) \times R_s + 2 \text{ V}$$ $$3 \text{ V} = (1.5 \text{ mA}) \times R_s$$ $$R_s = \frac{3 \text{ V}}{1.5 \text{ mA}} = 2 \text{ k}\Omega$$ This matches option 2. Note that if $R_s$ is less than $2 \text{ k}\Omega$, the current $I_{total}$ would increase, leading to a higher current through the LED and potentially exceeding its power rating. Thus, the minimum value for $R_s$ is $2 \text{ k}\Omega$.
