The Circuit Shown Has Four Identical Light Bulbs

The circuit shown has four identical light bulbs, and understanding how they behave together is a classic exercise in introductory electricity and magnetism. But by examining whether the bulbs are wired in series, parallel, or a combination of both, students can see how voltage, current, and resistance interact to determine brightness. This article walks through the reasoning step by step, provides the necessary formulas, and offers practical insights that help you predict what you would observe if you built the circuit in a lab.

Understanding the Basic Configuration

When you look at a diagram that contains four identical light bulbs, the first question to ask is: how are the bulbs connected? The answer determines everything that follows. In most textbook problems, the bulbs are arranged in one of three ways:

1. All four in series – a single continuous loop where the same current passes through each bulb.
2. All four in parallel – each bulb receives the full source voltage, but the total current splits among the branches.
3. A mixed arrangement – for example, two bulbs in series that are then placed in parallel with another pair of series bulbs.

Because the bulbs are identical, each has the same resistance, which we will denote as (R). But the power supply is assumed to be an ideal DC source with voltage (V_s). Knowing these two parameters lets us calculate the current through each bulb, the voltage drop across it, and ultimately its brightness, which is proportional to the power dissipated ((P = I^2R = V^2/R)) Simple as that..

It sounds simple, but the gap is usually here.

Series Connection: Same Current, Shared Voltage

Step‑by‑step Analysis

1. Total Resistance
   In a pure series circuit, resistances add:
   [ R_{\text{total}} = R + R + R + R = 4R ]

2. Circuit Current
   Using Ohm’s law ((I = V_s / R_{\text{total}})):
   [ I = \frac{V_s}{4R} ]
   This same current flows through every bulb because there is only one path.

3. Voltage Across Each Bulb
   The voltage drop across a single bulb is (V_{\text{bulb}} = I \times R):
   [ V_{\text{bulb}} = \frac{V_s}{4R} \times R = \frac{V_s}{4} ]
   Thus the source voltage is divided equally among the four bulbs Easy to understand, harder to ignore..

4. Power Dissipated (Brightness)
   [ P_{\text{bulb}} = I^2 R = \left(\frac{V_s}{4R}\right)^2 R = \frac{V_s^2}{16R} ]
   Alternatively, using (P = V_{\text{bulb}}^2 / R) gives the same result It's one of those things that adds up..

What You Would Observe

All four bulbs glow with identical brightness, but each is dimmer than a single bulb connected directly to the source (which would dissipate (V_s^2/R)). The series arrangement reduces the power per bulb by a factor of 1/16 compared to the solo‑bulb case Which is the point..

And yeah — that's actually more nuanced than it sounds Worth keeping that in mind..

Parallel Connection: Same Voltage, Shared Current

Step‑by‑step Analysis

1. Total Resistance
   For identical resistors in parallel, the equivalent resistance is:
   [ \frac{1}{R_{\text{total}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{4}{R} ]
   Hence
   [ R_{\text{total}} = \frac{R}{4} ]

2. Total Current from the Source
   [ I_{\text{total}} = \frac{V_s}{R_{\text{total}}} = \frac{V_s}{R/4} = \frac{4V_s}{R} ]

3. Current Through Each Bulb
   Because the voltage across each branch equals the source voltage, each bulb sees the full (V_s). The current through one bulb is:
   [ I_{\text{bulb}} = \frac{V_s}{R} ]
   Notice that the total current is four times this value, consistent with the parallel division Still holds up..

4. Power Dissipated (Brightness)
   [ P_{\text{bulb}} = \frac{V_s^2}{R} ]
   This is exactly the same power a single bulb would dissipate if it were the only load.

What You Would Observe

All four bulbs shine with the same brightness as a single bulb attached directly to the source. The parallel configuration does not dim any bulb; instead, it draws more total current from the supply.

Mixed Series‑Parallel Arrangement

A common textbook variant places two bulbs in series, then puts that series pair in parallel with another identical series pair. The diagram looks like two “branches,” each containing two bulbs end‑to‑end No workaround needed..

Step‑by‑step Analysis

1. Resistance of One Series Pair
   Two bulbs in series give (R_{\text{pair}} = R + R = 2R) That's the part that actually makes a difference..

2. Equivalent Resistance of Two Parallel Pairs
   Two identical branches of resistance (2R) in parallel yield:
   [ \frac{1}{R_{\text{total}}} = \frac{1}{2R} + \frac{1}{2R} = \frac{2}{2R} = \frac{1}{R} ]
   So (R_{\text{total}} = R) Most people skip this — try not to..

3. Source Current
   [ I_{\text{total}} = \frac{V_s}{R_{\text{total}}} = \frac{V_s}{R} ]

4. Current Through Each Branch
   Because the two branches are identical, the source current splits evenly:
   [ I_{\text{branch}} = \frac{I_{\text{total}}}{2} = \frac{V_s}{2R} ]

5. Current Through Each Bulb (within a branch)
   Inside a branch the bulbs are in series, so each bulb carries the branch current:
   [ I_{\text{bulb}} = \frac{V_s}{2R} ]

6. Voltage Across Each Bulb
   [ V_{\text{bulb}} = I_{\text{bulb}} \times R = \frac{V_s}{2R} \times R = \frac{V_s}{2} ]

7. Power Dissipated per Bulb
   [ P_{\text{bulb}} = I_{\text{bulb}}^2 R = \left(\frac{V_s}{2R}\right)^2 R = \frac{V_s^2}{4R} ]
   Or using (V_{\text{bulb}}^2 / R): ((V_s/2)^2 / R = V_s^2/(4R)).

What You Would Observe

Each bulb glows **