Question

Suppose we connect a 10-ohm light bulb to a 6-V battery. Some current flows through the battery (calculate it). Now, imagine that you add a 40-ohm resistor in parallel with the light bulb. By what factor does the current through the battery increase?

Answer 1

The current increases by the factor 1.25

Step-by-step explanation:

We have given that a 6 volt battery is connected to light bulb

So voltage V = 6 volt

Resistance connected R = 10 ohm

In first case

From ohm's law we know that current is given by

`i=(V)/(R)=(6)/(10)=0.6A`

In second case a 40 ohm resistor is connected in parallel

So equivalent resistance will be

`(1)/(R)=(1)/(10)+(1)/(40)`

`(1)/(R)=0.125`

R = 8 ohm

So the current will be
`i=(6)/(8)=0.75A`

So the factor by which current increases
`=(0.75)/(0.6)=1.25`

So the current increases by the factor 1.25