Question

(c) what is the resistance?

1. a student wishes to limit the current to an led (light - emitting diode) to 10.0 ma. the source voltage is +5 v and the diode drops 1.8 v.

(a) what value resistance is required?
(b) what power is dissipated in the resistor?

Explanation:

Step1: Calculate the voltage across the resistor

The source voltage $V_s = 5V$ and the diode - voltage drop $V_d=1.8V$. The voltage across the resistor $V_R$ is given by $V_R=V_s - V_d$. So, $V_R = 5 - 1.8=3.2V$.

Step2: Calculate the resistance value

We know from Ohm's law $V = IR$, where $I = 10.0mA=10\times10^{- 3}A$ and $V = V_R = 3.2V$. Rearranging for $R$, we get $R=\frac{V_R}{I}$. Substituting the values, $R=\frac{3.2}{10\times10^{-3}} = 320\Omega$.

Step3: Calculate the power dissipated in the resistor

The power dissipated in a resistor is given by $P = VI$. Here, $V = V_R = 3.2V$ and $I = 10\times10^{-3}A$. So, $P=3.2\times10\times10^{-3}=32\times10^{-3}W = 32mW$.

Answer:

(a) $320\Omega$
(b) $32mW$