Question
The maximum rated power of the LED is 2 mW and it is used in the circuit with input voltage of 5 V as shown in the figure below. The current through resistance $R_{S}$ is 0.5 mA .
The minimum value of the resistance of $R_{s}$, to ensure that the LED is not damaged is $\_\_\_\_$ $k\Omega$.
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Solution
Sol. [Key Concept: The LED-branch diode is reverse biased (no current there); the working bulb needs $P = VI$ so $V_{bulb} = 4$ V, leaving $V_{R_{s}} = 1$ V and $R_{s} = \frac{V_{R_{s}}}{I}$.]
Diode in reverse bias so no current in it.
$${P_{\text{bulb}\text{~}} = 2 \times 10^{- 3} = V_{\text{bulb}\text{~}} }{v_{\text{bulb}\text{~}} = 4\text{ }V }$$So $5\text{ }V = V_{RS} + V_{\text{bulb}\text{~}}$
$${V_{rs} = 5 - 4 = 1 }{1 = 0.5 \times 10^{- 3}R_{s} \Rightarrow R_{s} = 2k\Omega }$$
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