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Answer :
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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