Question

d) What will be the change in current if resistance is reduced to one-fourth of its original value?​

Answer 1

$\huge{\underline{\sf Lets \:Find:-}}$

Let

• Current=I
• Resistance=R
• Potential Difference=V

According to ohms law

$\boxed{\sf \dfrac{V}{I}=R}$

$\\ \sf \longmapsto I_1=\dfrac{V}{R}$

Now

• New Resistance=R/4

$\\ \sf \longmapsto \dfrac{V}{I}=\dfrac{R}{4}$

$\\ \sf \longmapsto 4V=IR$

$\\ \sf \longmapsto I_2=\dfrac{4V}{R}$

Change in current

$\\ \sf \longmapsto \dfrac{I_1}{I_2}=\dfrac{\dfrac{V}{R}}{\dfrac{4V}{R}}$

• Cancel V/R

$\\ \sf \longmapsto \dfrac{I_1}{I_2}=\dfrac{1}{4}$

$\\ \sf \longmapsto I_1:I_2=1:4$

Conclusion:-

• If resistance reduces to one fourth then Current through circuit increases four times than before.