Question 243: Current Electricity

Question

A conducting wire of length 'l', area of cross-section A and electric resistivity $$\rho$$ is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current.If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be :

Accepted Answer

We know that
$$R = ho {l \over A}$$
Now, new length : $$l' = 2l$$
new area of cross section : $$A' = A/2$$
$$ 	herefore $$ New resistance : $$R' = ho .{{2l} \over {A/2}}$$
$$ \Rightarrow R' = 4{{ho l} \over A}$$
$$ \Rightarrow R' = 4R$$
$$ 	herefore $$ Resultant current : $$I = {V \over {4R}}$$
$$ \Rightarrow $$ $$I = {1 \over 4}{{VA} \over {ho l}}$$

Answer

$$4{{VA} \over {\rho l}}$$

Answer

$${3 \over 4}{{VA} \over {\rho l}}$$

Answer

$${1 \over 4}{{VA} \over {\rho l}}$$

Answer

$${1 \over 4}{{\rho l} \over {VA}}$$

Current Electricity

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