Question

.A battery is connected to an electric lamp.If one more identical lamp is connected in series with the first what will happen to the current?​

Answer 1

An increase in the current is observed when the brightness of the bulb increases. If any series bulb is unscrewed, an open circuit is created and no current exists. Each device placed within the circuit provides a resistance to the electrons flowing through the wire.

Answer 2

The bridge you’re missing is resistance - particularly its relationship with the rest of Ohm’s law. Just knowing the equation means little: V=IR, or I=V/R right? Simple formula, but what does that actually mean?

V=IR: Voltage increase/decreases with Current and Resistance
I=V/R: Current and Resistance oppose each other
Plus you know how resistances work. Adding elements in parallel means the total resistance is less than any of its parts. You can see that easily with extremes:

With equal resistances, the resistance is cut in half -

1 ohm each: [1/x+1/x]^-1 = x/2
With drastically different resistances, the resistance is still less than the smaller one -

1 ohm and 10k ohms: [1/10000+1/1]^-1=0.999…
So to your question, you’re adding a resistance in parallel. You know now that obviously means the resistance is decreasing, no matter what it is. And what do we know about the relationship between resistance and current?

“I=V/R: Current and Resistance oppose each other”

When one goes up, one goes down. Simple see-saw logic!


See the image