Question

Question.2: Two heaters A and B are in parallel across supply voltage V. Heater A
produces 500 kcal in 20 min. and B produces 1000 kcal in 10 min. The resistance
of A is 10 ohm. What is the resistance of B ? If the same heaters are connected in
series across the voltage V, how much heat will be produced in kcal in 5 min?​

Answer 1

Answer:  2.5 ohm , 100 Kcal

Explanation:

First part:

We know that heat produced in a heater of resistance R and having voltage V in time t is given by,

$H = \frac{V^{2} }{R} t$

According to the question, voltage across the two heater A and B is same i.e. V . Let the resistance of the heater B is $R_B$ and that of A is $R_A = 10 ohm$.

so heat produced in A in time t is    $H_A = \frac{V^2}{R_A} t = 500 Kcal$

so heat produced in B in time t' is     $H_B = \frac{V^2}{R_B} t' = 1000 Kcal$

So we have,  [ t = 20 min, t' = 10 min ]

$\frac{H_A}{H_B} = 1/2\\\\\ or, \frac{R_B}{R_A}\frac{t}{t'} = 1/2\\\\or, R_B = \frac{1}{2} R_A\frac{t'}{t} \\ \\ = \frac{1}{2} \times 10 \times \frac{10}{20} ~~ohm \\ \\ = 2.5 ~~ohm$

From the first equation,  $V^2 = \frac{500 \times 10}{20}$  Kcal ohm/min

Second part:

When the two heaters are connected in series, let H be the amount of heat produced in Kcal. Since combined resistance is R = (10 + 2.5) = 12.5 ohm

When the heaters are connected in series, heat produced in time t = 5 min is,

$H = \frac{V^2}{R} t\\\\ = \frac{500 \times 10 }{20} \times 5 \times \frac{1}{12.5} ~~ Kcal\\\\= 100 ~~ Kcal$