Question

Q8. A resistor of length L is connected to a
battery and current I is flowing through it.
If it is divided into 3 parts by length and
all are having same cross sectional area are
connected in series with the same battery,
the current flowing through them will be:
a. 1/3 b. 31 c. 1 d. 31/2

Answer 1

Option c. is correct .

Given :

Resistor of length L .

Current flowing I .

Let , the area be A and resistance be R .

So , $R=\dfrac{\rho L}{A}$

Therefore , voltage applied is , $V=IR=\dfrac{I\rho L}{A}$ .

Now , resistance is divided into three parts of equal length i.e  $\dfrac{L}{3}$ and of same area A .

Therefore , new resistance is $R'=\dfrac{\rho L}{3A}$

Now , they are in series :

So , equivalent resistance , $R_{eq}=3R'=\dfrac{\rho L}{A}$

Now , New current , $I'=\dfrac{V}{R'}=\dfrac{\dfrac{I\rho L }{A}}{\dfrac\rho L{}{A}}= I$

Therefore , the new current is I .

Hence , this is the requires solution .

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Electric Current And Resistance

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Answer 2

Answer:

option c is correct bro

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