Series diode with resistor (Ohm’s law application): A diode is in series with a 220 Ω resistor. If the measured voltage across the resistor is 4 V, what is the current through the diode?

Introduction / Context:
In a simple series circuit, the same current flows through all series components. If you know the voltage across the series resistor, Ohm’s law directly gives the current in the entire branch, including the diode. This is a staple calculation in basic electronics troubleshooting and lab measurements.

Given Data / Assumptions:

• Series components: diode + 220 Ω resistor.
• Measured V across the resistor = 4 V.
• Ideal meter and steady-state conditions.

Concept / Approach:
Ohm’s law: I = V/R. In a series branch, current is identical through each element. Therefore, current through the resistor equals current through the diode. The diode’s forward drop is irrelevant here because we are not asked for supply voltage, only the branch current.

Step-by-Step Solution:
Use Ohm’s law: I = V/R.Substitute: I = 4 V / 220 Ω.Compute: 4 / 220 ≈ 0.01818 A.Convert to mA: ≈ 18.18 mA ≈ 18.2 mA.

Verification / Alternative check:
A quick mental check: 4/200 = 0.02 A → 20 mA; with 220 Ω it should be slightly less, about 18 mA—consistent with 18.2 mA.

Why Other Options Are Wrong:
(a) and (b) are unrealistically large by orders of magnitude. (d) 16.3 mA does not match 4/220 precisely. “None” is unnecessary since a precise match exists.

Common Pitfalls:
Using the diode drop to compute current even though the question already gives the resistor voltage; mixing up series versus parallel current rules.

Final Answer:
18.2 mA