The value of a series resistor required to limit the current t... | Technical Ability MCQ | Examoogle

Technical AbilityBasic Electrical Engineering

The value of a series resistor required to limit the current through an electric bulb to 40 mA with a forward voltage drop of 4V when connected to 16 V supply is ________.

$300 \Omega$

$20 \Omega$

$100 \Omega$

$1000 \Omega$

Correct Answer

⚙️ TE • Technical Concept & PrincipleTechnical AbilityBasic Electrical Engineering

Option A

$300 \Omega$

Quick Summary:

To limit the current through the bulb, a series resistor must drop the excess voltage from the source. The resistor acts as a current-limiting device governed by Ohm's Law.

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

To limit the current through the bulb, a series resistor must drop the excess voltage from the source. The resistor acts as a current-limiting device governed by Ohm's Law.

🔢 Key Formulas

$V_R = V_S - V_B$ — Voltage drop across the resistor

$R = \frac{V_R}{I}$ — Ohm's law to find the resistance

⚙️ Working Principle

The total supply voltage $V_S$ is shared between the bulb and the series resistor. The voltage across the resistor $V_R$ is calculated as the difference between the supply voltage and the bulb's forward voltage $V_B$ . Using Ohm's Law, the resistor value $R$ is determined by dividing $V_R$ by the desired current $I$ .

📌 Key Points

▸
The resistor must be connected in series to limit the current flowing through the circuit.
▸
The current $I$ remains constant through the series combination.
▸
The power rating of the resistor should also be considered in practical scenarios: $P = I^2 R$ .

✅ Advantages

▸
Simple and low-cost implementation
▸
Provides reliable current limiting for low-power LED or indicator bulb circuits

❌ Disadvantages / Limitations

▸
Power is wasted as heat in the resistor
▸
Voltage regulation is not as precise as an active current regulator

🛠️ Applications / Uses

▸
LED biasing circuits
▸
Indicator light current limiting

📄 Additional Information

▸
Calculation: $V_R = 16\text{ V} - 4\text{ V} = 12\text{ V}$ .
▸
Resistance $R = \frac{12\text{ V}}{40 \times 10^{-3}\text{ A}} = \frac{12}{0.04} = 300 \Omega$ .
▸
Option B (20 $\Omega$ ) represents $\frac{4}{0.04} - 80$ , while Option C (100 $\Omega$ ) results from miscalculating the voltage drop.

📊 Diagram / Illustration

✅

A is correct — The required series resistance is calculated as $300 \Omega$ to maintain the current at $40 \text{ mA}$ with a $12 \text{ V}$ drop.

Core Concepts Used

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Ohm's Law Kirchhoff's Voltage Law Series Circuit Analysis

💡 EXAM TIP

In competitive exams, always convert current from mA to A (divide by 1000) before applying Ohm's Law to avoid magnitude errors.

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