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Two heaters designed for the same voltag...

Two heaters designed for the same voltage `V` have different power ratings. When connected individually across as source of voltage` V`, they produce `H` amount of heat each in time `t_1` and `t_2` respectively. When used together acros the same source, they produce H amount of heat in time t

A

if they are in series `t=t_1+t_2`

B

if they are in series `t=2(t_1+t_2)`

C

if they are in parallel `t=(t_1t_2)/((t_1+t_2))`

D

if they are in parallel `t=(t_1t_2)/(2(t_1+t_2))`

Text Solution

Verified by Experts

The correct Answer is:
A, C
`H=V^2/R_1 t_1implies R_1=(V^2t_1)/H`
Similarly, `R_2=(V^2t_2)/H`
In series, `H=(V^2/(R_1+R_2))t`
`t=(H(R_1+R_2))/V^2`
substituting the value of `R_1` and `R_2` we get
`t=t_1+t_2`
In parallel, `H=V^2/R_("net")t=V^2t(1/R_1+1/R_2)`
`=V^2t(H/(V^2t_1)+H/(V^2t_2))`
Solving we get, `t=(t_1t_2)/(t_1+t_2)`
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Knowledge Check

  • The element of a heater is rated (P, V). If it is connected across a source of voltage V/2 , then the power consumed by it will be

    A
    P
    B
    2P
    C
    `P/2`
    D
    `P/4`

  • Two resistances , connected in parallel across a source of negligible interna resistance, consumes four times the power that they would consume when connected in series across the same source. If one of the resistances is 10 Omega the other is

    A
    `5 Omega`
    B
    `10 Omega`
    C
    `20 Omega`
    D
    `40 Omega`

  • Two electric bulbs A and B are designed for the same voltage. Their power ratings are P_A and P_B respectively with P_A lt P_B If they are joined in series across a V-volt supply:

    A
    A will draw more power than B
    B
    B will draw more power than A
    C
    A and B will draw the same power
    D
    nothing can be decided