Question

Average drift speed of charge carriers + ratios + resistances. Need help?

A `12V` battery with negligible internal resistance is connected to two resistors `Y` and `Z`, the resistors are connected in series.
The resistors are made from wires of the same material. The wire `Y` has a diameter `d` and length `l`. The wire `Z` has a diameter `2d` and length`2l`.

`(i)` Determine the ratio:
average drift speed of the charge carriers in `Y` `/` average drift speed of the charge carriers in `Z`.

`(ii)` Show that:
resistance of `Y` `/` resistance of `Z` `= 2`.

`(iii)` Determine the potential difference across `Y`.

`(iv)` Determine the ratio:
power dissipated in `Y` `/` power dissipated in `Z`.

P.S. sorry for long question... but it has to be done...

Answer 1

I got

Explanation:

`(i)` The average drift velocity `u` is given by the expression

`u = I /(n A q)`
where `I` is the current flowing in the resistor, `n` is the number density of charge-carrier, `A` is cross sectional area of the conductor, and `q` is the charge on the charge-carrier, in this case electronic charge.
`:.u_Y/u_Z=(I /(n A q))_Y/(I /(n A q))_Z`

As the resistors are made from the same wire `=>nand q` are same for both, are connected in series `=>` current flowing in the resistors is same. Above ratio reduces to

`u_Y/u_Z= A _Z/A_Y` .......(1)
`=>u_Y/u_Z= ((pid_Z^2)/4)/((pid_Y^2)/4)`

Inserting given values we get

`u_Y/u_Z= ((2d)^2)/d^2`
`=>u_Y/u_Z= 4`

`(ii)` Resistance `R` of a conducting wire is given as

`R=(rhoL)/A`
where `rho` is resistivity of material, `L` length of wire and `A` ist area of cross section.

Now required ratio

`R_Y/R_Z=((rhoL)/A)_Y/((rhoL)/A)_Z`

Noting that `rho` is same for both resistors,

`R_Y/R_Z=L_Y/L_Z A_Z/A_Y`

Using (1) and inserting given values we get

`R_Y/R_Z=l/(2l)xx 4`
`=>R_Y/R_Z=2`
Proved

`(iii)` Total resistance in the circuit

`R_T=R_Y+R_Z`
`=>R_T=R_Y+R_Y/2=3/2R_Y` .....(2)

From Ohm's law, Current in the circuit

`I=V/R_T`

Potential difference across resistor `Y` is give as

`V_Y=IR_Y`
`V_Y=(V/R_T)R_Y`

Using (2) we get

`V_Y=(V/(3/2R_Y))R_Y`
`=>V_Y=2/3V`

Inserting value of voltage source we get

`V_Y=2/3xx12=8V`

`(iv)` Power dissipated in resistor is given as

`P=I^2R`

The required ratio

`P_Y/P_Z=(I^2R)_Y/(I^2R)_Z`
`=>P_Y/P_Z=R_Y/R_Z=2`, (given in `(ii)` above)