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Power in Electric Circuits

For ensuring ...

Question

For ensuring dissipation of same energy in all three resistors $(R_{1}, R_{2}, R_{3})$ connected as shown in the figure, their values must be related as:

R_{1} = R_{2} = R_{3}

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R_{2} = R_{3}

and

R_{1} = 4 R_{2}

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R_{2} = R_{3}

and

R_{1} = (1 / 4) R_{2}

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R_{1} = R_{2} + R_{3}

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Solution

The correct option is B $R_{2} = R_{3}$ and $R_{1} = (1 / 4) R_{2}$
Let $i_{1}, i_{2} & i_{3}$ be current in $R_{1}, R_{2} & R_{3}$
Now, power consumed are same. So,
$i_{1}^{2} R_{1} = i_{2}^{2} R_{2} = i_{3}^{2} R_{3}$ ...(i)
And, $i_{2} R_{2} = i_{3} R_{3}$ [Potential drop over them are same] ...(ii)
Dividing (i) with (ii), $i_{2} = i_{3} \Rightarrow R_{2} = R_{3}$
$i_{1} = 2 i_{2}$ [ $i_{1}$ is being equally divided]
So, $i_{1}^{2} R_{1} = i_{2}^{2} R_{2} = \frac{i_{1}^{2}}{4} R_{2}$
or, $R_{2} = 4 R_{1}$

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Similar questions

(R_{1}, R_{2}, R_{3})

\frac{R_{1}}{r_{1}} + \frac{R_{2}}{r_{2}} + \frac{R_{3}}{r_{3}} =

Δ

r_{1} + r_{2} + r_{3} - r = 4 R

r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = Δ^{2}

△ A B C

a < b < c

r

r_{1}, r_{2}, r_{3}

A, B, C

r < r_{1} < r_{2} < r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

L - C - R

ω

R_{1}, R_{2} a n d R_{3}

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