If power developed across 6 - resistor is equal to the power developed across 24 - resistor- The value of R isA- 12 -B- 6 -C- 24 -D- 8 -

The correct option is C 24ΩWe have, I_1=RI/R+24 I_2=24I/R+24 P_1=R^2 I^2/(R+24)^2× 24 P_2=I^2.6 P^2=P_1 ∴ R=24Ω. ...

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Standard XII

Physics

Motional EMF in an Open Conductor

If power deve...

Question

If power developed across $6 Ω$ resistor is equal to the power developed across $24 Ω$ resistor. The value of R is

12 Ω

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6 Ω

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24 Ω

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8 Ω

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Solution

The correct option is C $24 Ω$
We have, $I_{1} = \frac{R I}{R + 24}$ $I_{2} = \frac{24 I}{R + 24}$
$P_{1} = \frac{R^{2} I^{2}}{(R + 24)^{2}} \times 24$ $P_{2} = I^{2} .6 P^{2} = P_{1}$
$∴ R = 24 Ω$ .

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If power developed across 6Ω resistor is equal to the power developed across 24Ω resistor. The value of R is

The correct option is C 24ΩWe have, I1=RIR+24 I2=24IR+24 P1=R2I2(R+24)2×24 P2=I2.6 P2=P1 ∴R=24Ω.

If power developed across $6 Ω$ resistor is equal to the power developed across $24 Ω$ resistor. The value of R is

The correct option is C $24 Ω$
We have, $I_{1} = \frac{R I}{R + 24}$ $I_{2} = \frac{24 I}{R + 24}$
$P_{1} = \frac{R^{2} I^{2}}{(R + 24)^{2}} \times 24$ $P_{2} = I^{2} .6 P^{2} = P_{1}$
$∴ R = 24 Ω$ .