To boil a certain mass of water, a coil will take a time of $t_{1}$ and another coil will take a time of $t_{2}$ , what will be the time taken when coils are connected in (a) series (b) parallel

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Solution

Let the resistance of the coils are R₁ and R₂

Heat produced in coil 1 $H_{1} = \frac{{V_{1}}^{2}}{R_{1}} t_{1}$

Heat produced in coil 2 $H_{2} = \frac{{V_{2}}^{2}}{R_{2}} t_{2}$

When the two coils are connected in series, current is same

$R_{e q} = R_{1} + R_{2}$

$R_{e q} = \frac{{V_{1}}^{2}}{H_{1}} t_{1} + \frac{{V_{2}}^{2}}{H_{2}} t_{2}$

$\frac{I^{2} R^{2}}{H} t = \frac{I^{2} {R_{1}}^{2}}{H_{1}} t_{1} + \frac{I^{2} {R_{2}}^{2}}{H_{2}} t_{2}$

$t = \frac{H}{R^{2}} (\frac{{R_{1}}^{2}}{H_{1}} t_{1} + \frac{{R_{2}}^{2}}{H_{2}} t_{2})$

When the two coils are connected in parallel, voltage is same

$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$\frac{1}{\frac{V^{2}}{H} t} = \frac{1}{\frac{V^{2}}{H_{1}} t_{1}} + \frac{1}{\frac{V^{2}}{H_{2}} t_{2}}$

$t = \frac{H}{V^{2} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{1}{\frac{V^{2}}{H_{1}} t_{1}} + \frac{1}{\frac{V^{2}}{H_{2}} t_{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠}$

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