
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
232.8k+ 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
Circuit Switching vs Packet Switching: Key Differences Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Electricity and Magnetism Explained: Key Concepts & Applications

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

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

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

Understanding Uniform Acceleration in Physics

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

