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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?

Respuesta :

Answer:

The current increases by the factor 1.25                  

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

[tex]i=\frac{V}{R}=\frac{6}{10}=0.6A[/tex]

In second case a 40 ohm resistor is connected in parallel

So equivalent resistance will be

[tex]\frac{1}{R}=\frac{1}{10}+\frac{1}{40}[/tex]

[tex]\frac{1}{R}=0.125[/tex]

R = 8 ohm

So the current will be [tex]i=\frac{6}{8}=0.75A[/tex]

So the factor by which current increases [tex]=\frac{0.75}{0.6}=1.25[/tex]

So the current increases by the factor 1.25  

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