Question

You have 10 ohm and a 100 ohm resistor in parallel. You place this equivalent resistance in series with an LED, which is rated to have a voltage drop of 1.83V at 20mA. What voltage should you connect to this circuit to provide 20mA of current for the LED? What would the current be, if you connected to a 3V battery? Draw all of these diagrams before attempting the calculations.

Answer 1

Answer:

Approximately $\rm 2.0\; V$.

Approximately $\rm 30 \; mA$. (assumption: the LED here is an Ohmic resistor.)

Explanation:

The two resistors here $R_1= 10\; \Omega$ and $R_2= 100\; \Omega$ are connected in parallel. Their effective resistance would be equal to

$\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega$.

The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be $20\; \rm mA = 0.020\; \rm A$. That should also be the current through the effective $\displaystyle \rm \frac{10}{11} \; \Omega$ resistor. Make sure all values are in standard units. The voltage drop across that resistor would be

$V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V$.

The voltage drop across the entire circuit would equal to

• the voltage drop across the resistors, plus
• the voltage drop across the LED.

In this case, that value would be equal to $1.83 + 0.182 \approx 2.0\; \rm V$. That's the voltage that needs to be supplied to the circuit to achieve a current of $20\; \rm mA$ through the LED.

Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:

$\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega$.

The resistance of a serial circuit is equal to the resistance of its parts. In this case,

$\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega$.

Again, the current in a serial circuit is the same in all appliances.

$\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA$.