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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityEquivalent ResistanceMedium2 minPYQ_2012

PhysicsMediumsingle choice

The power dissipated in the circuit shown in the figure is 30 Watt. The value of $R$ is
NEET 2012 (Mains)

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Answer:

Solution:

Here, $R_{1} = R =$ ? $R_{2} = 5Ω, V = 10 V$ and $P = 30 W$ Hence $P \frac{1 0 ^{2}}{R} \frac{100}{R} R = \frac{V ^{2}}{R _{1}} + \frac{V ^{2}}{R _{2}} = 30 - \frac{1 0 ^{2}}{5} = 30 - 20 = 10Ω$

Stream:NEETSubject:PhysicsTopic:Current ElectricitySubtopic:Equivalent Resistance

⏱ 2mℹ️ Source: PYQ_2012

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