Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityElectrical PowerMedium2 minQB

PhysicsMediumsingle choice

The resistive network shown below is connected to a $D . C .$ source of $16 V .$ The power consumed by the network is $4$ Watt. The value of $R$ is:

Options:

Answer:

Solution:

$R_{e q}$ for mixed series and parallel combination of

given circuit $R_{e q} = \frac{1}{\frac{1}{4 R} + \frac{1}{4 R}} + R + \frac{1}{\frac{1}{6 R} + \frac{1}{12 R}} + R$

$R_{e q} = 8 R$
Power dissipated by circuit $P = \frac{V ^{2}}{R _{e q}}$
$\Rightarrow 4 = \frac{1 6 ^{2}}{8 R}$
$\Rightarrow R = \frac{256}{32} = 8Ω$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Electrical Power

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...