
A simple circuit consists of a battery and a light bulb. A second identical light bulb is then added in series to the first light bulb. What effect does this have on the brightness of the first light bulb?
A) The brightness gets brighter
B) The light bulb goes out
C) The brightness gets dimmer
D) The brightness does not change
Answer
235.5k+ views
Hint:-The brightness of a bulb depends upon the power dissipated by the bulb. We just have to compare power dissipation of the bulb in initial condition with the final condition after adding another bulb in series with it.
Complete Step by Step Explanation:
When we add a resistance in series with an already present resistance in a circuit, then the initial voltage difference across the initial resistance is divided among both resistances after adding them in series according to their resistance values.
This is due to decrease in the current, because according to ohm’s law,
$V = IR$
Where $R$ is the resistance
$I$ is the current passing through $R$
And $V$ is the voltage across $R$ .
Now, if we add another resistance in series with $R$ , equivalent resistance of the circuit will increase, which states that $I$ will decrease, because $V$ will remain constant across the circuit.
Now we know that power $P = \dfrac{{{V^2}}}{R}$
Now, putting value $V = IR$ (by ohm’s law) in above equation, we get,
$P = \dfrac{{{I^2}{R^2}}}{R}$
On solving, we get,
$P = {I^2}R$
Therefore, $P \propto {I^2}$
Now, the total resistance of the circuit is increasing due to addition of another bulb in series with the first bulb of the circuit when there was no second bulb, therefore the current is decreasing due to increase in resistance, so power dissipation will also decrease in the first bulb.
So, the brightness of the bulb will decrease.
So, the correct answer is option (C).
Note: Since, we added the bulb in series, therefore the current was decreasing and voltage was constant, but if resistance would be added in parallel, then resistance would have been decreased so current would increase, and brightness would increase.
Complete Step by Step Explanation:
When we add a resistance in series with an already present resistance in a circuit, then the initial voltage difference across the initial resistance is divided among both resistances after adding them in series according to their resistance values.
This is due to decrease in the current, because according to ohm’s law,
$V = IR$
Where $R$ is the resistance
$I$ is the current passing through $R$
And $V$ is the voltage across $R$ .
Now, if we add another resistance in series with $R$ , equivalent resistance of the circuit will increase, which states that $I$ will decrease, because $V$ will remain constant across the circuit.
Now we know that power $P = \dfrac{{{V^2}}}{R}$
Now, putting value $V = IR$ (by ohm’s law) in above equation, we get,
$P = \dfrac{{{I^2}{R^2}}}{R}$
On solving, we get,
$P = {I^2}R$
Therefore, $P \propto {I^2}$
Now, the total resistance of the circuit is increasing due to addition of another bulb in series with the first bulb of the circuit when there was no second bulb, therefore the current is decreasing due to increase in resistance, so power dissipation will also decrease in the first bulb.
So, the brightness of the bulb will decrease.
So, the correct answer is option (C).
Note: Since, we added the bulb in series, therefore the current was decreasing and voltage was constant, but if resistance would be added in parallel, then resistance would have been decreased so current would increase, and brightness would increase.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Young’s Double Slit Experiment Derivation Explained

Wheatstone Bridge – Principle, Formula, Diagram & Applications

Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Electromagnetic Waves and Their Importance

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Essential Derivations for CBSE Class 12 Physics: Stepwise & PDF Solutions

CBSE Class 12 Physics Question Paper Set 1 (55/1/1) 2025 – PDF, Solutions & Marking Scheme

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Understanding Uniform Acceleration in Physics

