(a) When a cell sends current through a resistance `R_(1)` for time `t`, the heat produced in the resistor is `Q`. When the same cell sends current through another resistance `R_(2)` for time `t` the heat porduced is again `Q` . Find internal resistance of the cell.
(b) An electric kettle has coil `A` is swithched on, the water boils in 10min. and when coil b is switched on the water boils in `20min`. Calculate the time taken by the water to boil if the coils connected in (i) series and (ii) parallel are swithched on.
(c) The emf of battery `E` in the circuit shown in the figure is 15V and internal resistance `0.5ohm` (i) What is the current drown from the battery? (ii) How much power is consumed in `6 ohm` resistance?

(d) A `500W` heating unit is designed to opreate from a `115V` line.
(i) By what percentage will its heat output drop if the line voltage drops to `110V`. Assuming no change in resistance?
(ii) Taking variation of resistance with temperature into account. would the actual heat output drop be larger or smaller than case(i)
(e) If emf of the cell is `1.8V` and its internal resistance `2//3 Omega` for the circuit shown in the figure what is the (i) current in `3 ohm` resistance? (ii) Power consumed by the circuit from the battery?

(i) for what value of `R` , power in `R` is maximum?
.

Explanation

(a)
`i_(1) - (V)/(R_(1)+r)`
`H_(1) = i_(1)^(2)R_(1)t = ((V)/(R_(1)+r))^(2) R_(1)t` ..(i)

`i_(2) - (V)/(R_(2)+r)`
`H_(2) = i_(2)^(2)R_(2)t = ((V)/(R_(2)+r))^(2) R_(2)t` ..(ii )
(i)=(ii)
` ((V)/(R_(1)+r))^(2) R_(1)t=((V)/(R_(2)+r))^(2) R_(2)t`
`(R_1)/((R_(1)+r)^(2))=(R_3)/((R_(2)+r)^(2))`
`R_(2)^(2)R_(1) +2R_(2)r R_(1)+r^(2)R_(1)=R_1^2R_(2)+2R_1rR_2+r^2R_2`
`r^2(R_1-R_2)=R_1R_2(R_1-R_2)`
`r=sqrt(R_1R_2)`

`H = (V^2)/(R_2)t_2=(V^2)/(R_2)t_2= (V^2)/(R_1+R_2)t_(1)^(') = (V^2)/((R_1R_(2))/(R_1+R_2)t_2^'`
`(V^2)/(R_1)t_1=(V^2)/(R_2)t_2-R_2=(t_2)/(t_1)R_1`
`(V^2)/(R_1)t_1 = (V^2)/(R_1+R_2)t_1^' = (V^2)/(R_1_(t_2)/(t_1)R_(1)) t_1^'`
`t_(1) = (t_1^')/(1+(t_2)/(t_1) implies t_(1)^' = t_1+t_2=10+20=30min`
`(V^2)/(R_1)t_1 = (V^2)/((R_1R_2)/(R_1+R_2))t_2^'implies t_1 = (R_1+R_2)/(R_2) t_(2)^(')`
`((R_1)/((t_2//t_1)R_1)+1)t_2^'`
`t_(1)=((t_1)/(t_t2)+1)t_2^' implies 1/(t_2^') = 1/(t_1)+1/(t_2)+1/10+1/20`
`t_(2)^(') = 20/3 min`
(c)
(i) `R_(eq) = 15 Omega`
Current supplied by battery
`i = (15)/(15) = 1A`
p.d. across `4.5 Omega`
`=1xx4.5=4.5 V`
(ii) Power consumed in `6 Omega`
`=((4.5)^(2))/(6) = 3.375W`
(d) (i) `P=(V^2)/R`
`(P_2)/(P_1)=((V_2)/(V_1))^(2) = (110/115)^(2) = (22/23)^(2) = ((23-1)/(23))^(2)`
`=(1-1/23)^(2) = 1-2/23`
`((P_2)/(P_1)-1) xx 100 = - 200/23 = -8.7%`
ie., `8.7%` decrease.
(ii) As temperature increases, resistance increases
`P = (V^2)/(R)`
Power output will be smaller.

`i=1.8/(4/3+2/3)=0.9A`
p.d. across `4/3Omega`
`V' = ixx4/3=0.9xx4/3=1.2V`
Current in `3Omega i_(0)=(V')/(3)=1.2/3 = 0.4A`
Power consumed in circuit - `i^(2)R_(eq)`
`=(0.9)^(2)(4/3+2/3) = 1.62W`
(f)
`i_(1) = (V)/((30R)/(30+R)+20)=(V(30+R))/((600+50R))`
Power in `R`
`P = i_(1)^(2)R = [(3V)/(60+5R)]^(2) R`
`=9V^(2) = R/((60+5R)^(2))`
For `P` to be maximum
`(dP)/(dR) = 9V^(2)((60+5R)^(2)*1-R*2(60+5R)*5)/((60+5R)^(4))`
`(60+5R) - 10R = 0R = 12 Omega`.