Two condenser of 2 - F and 4 - F are connected in series- The p-d of 1200 volt- The p-d across 2 - F is -

The correct option is C 800 V REF.ImageQ=CVFor a constant charge , V ∝1/C V_A∝1/C_A #160; #160; amp; #160; V_B∝1/C_B V_A=CB/C ...

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Byju's Answer

Standard XII

Physics

Electric Potential at a Point in Space Due to a Point Charge Q

Two condenser...

Question

Two condenser of $2 μ F$ and $4 μ F$ are connected in series. The $p . d$ of $1200$ volt. The $p . d$ across $2 μ F$ is -

400 V

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600 V

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800 V

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900 V

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Solution

The correct option is C $800 V$
REF.Image
Q=CV
For a constant charge , $V \propto \frac{1}{C}$
$V_{A} \propto \frac{1}{C_{A}}$ & $V_{B} \propto \frac{1}{C_{B}}$
$V_{A} = \frac{C B}{C A + C B} \times V$
$V_{A} = (\frac{4}{2 + 4}) \times 1200$
$= \frac{4}{6} \times 1200$
= 800 V

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Two condenser of 2μF and 4μF are connected in series. The p.d of 1200 volt. The p.d across 2μF is -

The correct option is C 800 VREF.ImageQ=CVFor a constant charge , V∝1CVA∝1CA & VB∝1CBVA=CBCA+CB×VVA=(42+4)×1200=46×1200= 800 V

Two condenser of $2 μ F$ and $4 μ F$ are connected in series. The $p . d$ of $1200$ volt. The $p . d$ across $2 μ F$ is -

The correct option is C $800 V$
REF.Image
Q=CV
For a constant charge , $V \propto \frac{1}{C}$
$V_{A} \propto \frac{1}{C_{A}}$ & $V_{B} \propto \frac{1}{C_{B}}$
$V_{A} = \frac{C B}{C A + C B} \times V$
$V_{A} = (\frac{4}{2 + 4}) \times 1200$
$= \frac{4}{6} \times 1200$
= 800 V