For the given circuit, three resistances R, 2R and 6R are given. Ratio of power developed across R, 2R and 6R will be

! : 2 : 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 : 4 : 3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 : 5 : 8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 : 3 : 5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 : 4 : 3$
The current divides as i and $\frac{i}{2}$
Power $P = i^{2} R$
$∴ P_{R} = i^{2} R$
$P_{2 R} = i^{2} (2 R)$
$P_{6 R} = {(\frac{i}{2})}^{2} (6 R)$
$∴ P_{R} : P_{2 R} : P_{6 R} = 1 : 2 : 1.5 = 2 : 4 : 3$

Suggest Corrections

Similar questions

10 \sum r = 1 r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)

α

In the circuit given, power developed across 1Ω,2Ω and 3Ω resistances are in the ratio of

r_{+}

r_{-}

3

4

\frac{3}{4}

R

Join BYJU'S Learning Program