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Standard VIII

Physics

Heating Effect of Electric Current

In the circui...

Question

In the circuit shown, the resistance $r$ is a variable resistance. If for $r = f R$ , the heat generation in $r$ is maximum then the value of $f$ is:

1

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\frac{1}{4}

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\frac{1}{2}

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\frac{3}{4}

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Solution

The correct option is C $\frac{1}{2}$
Heat generation in r will be, $H =$ $I_{r}^{2} r$ ,

$H = {(\frac{V}{R + \frac{R r}{R + r}} \times \frac{r}{R + r})}^{2} r = {(\frac{V}{R (R + 2 r)})}^{2} r$

Putting $\frac{d H}{d r} = 0$ gives, $r = \frac{R}{2}$ . Hence, $f = \frac{1}{2}$ .

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In the circuit shown, the resistance r is a variable resistance. If for r=fR, the heat generation in r is maximum then the value of f is:

The correct option is C 12Heat generation in r will be, H= I2rr,H=(VR+RrR+r×rR+r)2r=(VR(R+2r))2rPutting dHdr=0 gives, r=R2. Hence, f=12.

In the circuit shown, the resistance $r$ is a variable resistance. If for $r = f R$ , the heat generation in $r$ is maximum then the value of $f$ is:

The correct option is C $\frac{1}{2}$
Heat generation in r will be, $H =$ $I_{r}^{2} r$ ,

$H = {(\frac{V}{R + \frac{R r}{R + r}} \times \frac{r}{R + r})}^{2} r = {(\frac{V}{R (R + 2 r)})}^{2} r$

Putting $\frac{d H}{d r} = 0$ gives, $r = \frac{R}{2}$ . Hence, $f = \frac{1}{2}$ .