Solve this---**Question Stem:**
WHICH LIGHT BULB WILL GLOW BRIGHTER?
**Chart/Diagram Description:**
* **Type:** Electrical circuit diagram showing two light bulbs connected in series to an AC power source.
* **Main Elements:**
* Two identical-looking incandescent light bulbs are shown.
* The left bulb is labeled "60W".
* The right bulb is labeled "100W".
* The bulbs are connected in series via black lines representing wires.
* The circuit is connected to a power source labeled "L1" and "N".
* Between L1 and N, the voltage is indicated as "~220V". The "~" symbol indicates AC voltage.
* The connection from L1 goes to the bottom terminal of the left bulb.
* The top terminal of the left bulb is connected to the top terminal of the right bulb.
* The bottom terminal of the right bulb is connected to N.
* **Labels and Annotations:**
* "WHICH LIGHT BULB WILL GLOW BRIGHTER?" (Question title)
* "60W" (Label for the left bulb)
* "100W" (Label for the right bulb)
* "L1" (Label for one terminal of the power source, colored red)
* "~220V" (Voltage and type of power source)
* "N" (Label for the other terminal of the power source, colored blue)
* "WORLD OF ENGINEERING" (Text next to a gear logo at the bottom right)
**Other Relevant Text:**
WORLD OF ENGINEERING
视频信息
答案文本
视频字幕
We have a circuit with two light bulbs connected in series. A 60-watt bulb and a 100-watt bulb are connected to a 220-volt AC power source. The question is: which bulb will glow brighter? In a series circuit, the same current flows through both bulbs.
To solve this problem, we need to calculate the resistance of each bulb using the power formula. The resistance equals voltage squared divided by power. For the 60-watt bulb, the resistance is 220 squared divided by 60, which equals 806.67 ohms. For the 100-watt bulb, it's 220 squared divided by 100, which equals 484 ohms. The 60-watt bulb has higher resistance.
In a series circuit, the same current flows through both bulbs. The power dissipated by each bulb is calculated using P equals I squared times R. Since the current is the same for both bulbs, the power depends only on resistance. The 60-watt bulb with higher resistance will dissipate more power than the 100-watt bulb. Higher power dissipation means brighter light, so the 60-watt bulb will glow brighter.
Let's calculate the exact values. The total resistance is 806.67 plus 484 equals 1290.67 ohms. The current is 220 volts divided by 1290.67 ohms, which equals 0.170 amperes. The actual power dissipated by the 60-watt bulb is 0.170 squared times 806.67, which equals 23.3 watts. The 100-watt bulb dissipates 0.170 squared times 484, which equals 14.0 watts. Therefore, the 60-watt bulb glows brighter with 23.3 watts compared to the 100-watt bulb's 14.0 watts.
Let's summarize the key concepts. Power ratings are given for normal operating voltage. In a series circuit, the same current flows through all components. The component with higher resistance dissipates more power, and more power means brighter light. This is why the 60-watt bulb glows brighter than the 100-watt bulb when connected in series. In normal parallel operation, the 100-watt bulb would be brighter, but series connection reverses this relationship due to the resistance differences.