A 110 , V d.c. heater is used on an a.c. sour | vedclass

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Question

(A) The heat produced by a heater in a resistance $R$ over time $t$ is given by $H = \frac{V^2}{R} t$.For $d.c.$ voltage,$H_{dc} = \frac{V_{dc}^2}{R}

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$110$

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(A) The heat produced by a heater in a resistance $R$ over time $t$ is given by $H = \frac{V^2}{R} t$.For $d.c.$ voltage,$H_{dc} = \frac{V_{dc}^2}{R} t$.For $a.c.$ voltage,the heat produced is determined by the $r.m.s.$ value of the voltage,$H_{ac} = \frac{V_{rms}^2}{R} t$.Given that the heat produced is the same in both cases $(H_{ac} = H_{dc})$,we have:$\frac{V_{rms}^2}{R} t = \frac{V_{dc}^2}{R} t$$V_{rms}^2 = V_{dc}^2$$V_{rms} = V_{dc}$Given $V_{dc} = 110 \, V$,therefore $V_{rms} = 110 \, V$.

$A$ $110 \, V$ $d.c.$ heater is used on an $a.c.$ source,such that the heat produced is the same as it produces when connected to $110 \, V$ $d.c.$ in the same time intervals. What would be the $r.m.s.$ value of the alternating voltage?

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