Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

Two capacitors of capacity $6 μ F$ and $12 μ F$ in series are connected by potential of $150 V$ . The potential of the capacitor of capacity $12 μ F$ will be

Options:

Answer:

Solution:

Given, $C_{1} = 6 μ F, C_{2} = 12 μ F, V = 150 v o l t$ .

Total capacity, $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{6} + \frac{1}{12}$

$\frac{1}{C} = \frac{3}{12} \Rightarrow C = 4 μ F$

Potential of $12 μ F$ capacitor

$V = \frac{q}{C}$

$V = \frac{4 \times 150}{12}$

$V = 50 V$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...