Question

3. Imagine a simple series circuit with one 1.5V battery and one bulb. When the
1.5V battery is replaced with a 3V battery ...
a) the bulb gets brighter
b) the bulb gets dimmer
c) the bulb stays at the same level of brightness

Answer 1

Answer:

The correct option is (a) The bulb gets brighter.

Explanation:

The brightness of a bulb depends upon the voltage supplied and current. The current in turn depends on the voltage too. This means if the voltage is increased the bulb gets brighter and the current increase. When the voltage is decreased, the bulb gets dimmer and a low current is produced.

Conclusion:

When a 3V battery is used in place of a 1.5V battery, the bulb gets brighter.

Answer 2

When the 1.5V battery is replaced with a 3V battery, the bulb gets brighter (Option - a)

• The battery is always connected parallel to the bulb.
• The intensity of light emitted by the bulb or the brightness of the bulb depends on the amount of current passing through the bulb.
• Earlier the battery connected to the circuit had a voltage of 1.5V, replaced by a higher voltage battery of 3V.
• Current passing through the bulb is calculated by ohm's law i.e. I = V/R (V - voltage, I - current, R - Resistance of the bulb).
• Current passing through the bulb is directly proportional to the voltage applied. Hence, the higher the voltage difference -the higher is the current passing through the bulb and the higher is its brightness.
• 3V > 1.5V hence more amount of current passes through the bulb and it glows brighter.