Question

A particular resistance wire has a resistance o
3.0 ohm per metre. Find :
(a) The total resistance of three lengths of this
wire each 1.5 m long, joined in parallel.
(b) The potential difference of the battery whid
gives a current of 2.0 A in each of the 1.5
length when connected in parallel to the battery
(assume that the resistance of battery is
negligible).​

Answer 1

Answer:

because the battery cannot flow upwards

Explanation:

Answer 2

Answer:

$\LARGE{ \underline{ \bf{Required \: answer:}}}$

$\large{ \boxed{ \boxed{ \rm{1.5 \: ohm}}}}$

Explanation:

We have,

Resistance of 1 m wire = 3 Ω

Now, we know that resistance is directly proportional to length and is given by R = Resistivity × Length / Cross-sectional Area. Resistivity and Cross sectional area will remain same, then Resistance will vary with length.

Then,

Resistance of 1.5 m wire = 3 × 1.5 Ω = 4.5 Ω

So, we have three lengths of resistors with resistance 4.5 Ω. They are connected in parallel. Then equivalent resistance:

$\rm{ \dfrac{1}{R_eq.} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}$

We have, R1 = R2 = R3. Hence Req. = R/3

$\rm{R_eq. = \dfrac{4.5}{3} } = 1.5 ohm$

And we are done! :D