Four equal resistance dissipated $5 W$ of power together when connected in series to a battery of negligible internal resistance. The total power dissipated in these resistance when connected in parallel across the same battery would be

125 W

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

80 W

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

20 W

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

5 W

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

Open in App

Solution

The correct option is B $80 W$
In series, $R_{s} = 4 R$
Power $P = \frac{V^{2}}{4 R}$
$⟹ 5 = \frac{V^{2}}{4 R}$
$⟹ 20 = \frac{V^{2}}{R} \to (1)$
In parallel, $R + p = \frac{R}{4}$
Power consumed, $P . \frac{v^{2}}{R_{p}} = 4 \frac{v^{2}}{R}$
$4 \times 20$
$= 80 W$

Suggest Corrections

Similar questions

The equal resistors connected in series across a source of emf together dissipate 5W of power. Of the same resistors are connected in parallel across the same source of emf , the power dissipated will be?

Q. Four 2 ohm resistors are connected together along the edges of a square. A 10V battery of negligible internal resistance is connected across a pair of the diagonally opposite corners of the square. The power dissipated in the circuit is

Two resistors A and B resistance 4 $Ω a n d 6 Ω$ respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate : (i) the power supplied by the battery, and (ii) the power dissipated in each resistor.

Q. The power dissipated by a light bulb with 4 ohm resistance when connected in parallel to 12V battery is

Q. Two equal resistance, when connected in series to a battery, consume electric power of

60 W

. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be:

Join BYJU'S Learning Program

Four equal resistance dissipated 5W of power together when connected in series to a battery of negligible internal resistance. The total power dissipated in these resistance when connected in parallel across the same battery would be

The correct option is B 80WIn series,Rs=4RPower P=V24R⟹5=V24R⟹20=V2R→(1)In parallel, R+p=R4Power consumed, P.v2Rp=4v2R4×20=80W

Four equal resistance dissipated $5 W$ of power together when connected in series to a battery of negligible internal resistance. The total power dissipated in these resistance when connected in parallel across the same battery would be

The correct option is B $80 W$
In series, $R_{s} = 4 R$
Power $P = \frac{V^{2}}{4 R}$
$⟹ 5 = \frac{V^{2}}{4 R}$
$⟹ 20 = \frac{V^{2}}{R} \to (1)$
In parallel, $R + p = \frac{R}{4}$
Power consumed, $P . \frac{v^{2}}{R_{p}} = 4 \frac{v^{2}}{R}$
$4 \times 20$
$= 80 W$