(a) Two heater coils made of the same material are connected in paralled across the mains. The length of diameter of one the coil is triple that of other. Which of these will produce more heat?
(b) Three equal resistors connected in series across a source of emf together dissipated `10 Watt` of power. What would be the power disspated if the same resistors are connected in parallel across the same source of emf?
Two resistor `R_(1)` and `R_(2)` may be connect either in series or in parellel across a battery of zero internal resistance. it is required that joule heating for the parallel combination be four times than for series combination. if `R_(1)= 100 Omega` find `R_(2)`
(d) The three resistance `A, B and C` have values `3R, 6R` and `R` respectively. Now some potential difference is applied across the network . Find the ratio of thermal powers dissipated between `X` and `Y` . find ratio of power consumed by `A,B, and C`.

Explanation

(a) `R_(1) = (rhol)/((pid^(2)//4)),R_(2)=(rho.3l)/(pi/4 (3d)^(2))=1/3(rhol)/((pid^(2)//4))=(R_1)/3`
Heat produced in coils
Coil `1, H_(1) = (V^2)/(R_1)t`
Coil2, `H_(2)= (V^2)/(R^2) t`
`(H_2)/(H_1) = (R_1)/(R_2) = (R_1)/(R_(1)//3) = 3`
`H_(2) gt H_(1)` second coil
(b)
`P=(V^2)/(R_(eq)) = (V^2)/(3R)`
`10=(V^2)/(3R)`
`(V^2)/(R) = 30`

`P' = (V^2)/(R_(eq)) = (V^2)/(R//3) = (3V^2)/(R)`
`=3xx30=90W`
(c)
`P_(1)=(V^2)/(R_(1)+R_(2))`

`P_(2) = (V^2)/(R_(1))R_(2)//(R_(1)+R_(2)))`
`P_(2)=4P_(1)`
`(V^2(R_(1)+R_(2))/(R_(1)R_(2)) = 4(V^2)/((R_(1)+R_(2))`
`(R_(1)+R_(2))^(2) = 4R_(1)R_(2)`
`R_(1)^(2) - 2 R_(1)R_(2) +R_(2) = 0`
`(R_(1)-R_(2))^(2)=0 implies R_(2) = R_(1) = 100 Omega`
(d)
`i_(1) = (6R)/(3R+6R)i=(2i)/(3)`
`i_(2_) = i-i_(1) = i/3`
`P_(A) : P_(B):P_(C) = ((2i)/(3))^(2) . 3R : (i/3)^(2) . 6R : i^(2)R`
`=4/3:2/3:1`
`=4:2:3:`