Two condensers of capacitance 4mu F and 5mu F are joined in series- If the potential difference across 5mu F is 10V- then the potential difference across 4mu F condenser is -22-5V10V12-5V25V

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Toppr Admin · Accepted Answer

When two capacitors are in series Q present on then is same-x2234-xA0-Q-xA0-on-xA0-5-x3BC-F is Q-5-xD7-10-x2212-6-xD7-10Q-5-xD7-10-x2212-5-x2234-Q on 4-x3BC-F is 5-xD7-10-x2212-5-x2234-V-QCV-5-xD7-10-x2212-54-xD7-10-x2212-6V-12-5-xA0-V

Two condensers of capacitance $4 μ F$ and $5 μ F$ are joined in series. If the potential difference across $5 μ F$ is $10 V$ , then the potential difference across $4 μ F$ condenser is :

$22.5 V$
$10 V$
$12.5 V$
$25 V$

When two capacitors are in series $Q$ present on then is same
$∴ Q o n 5 μ F$ is $Q = 5 \times 10^{- 6} \times 10$
$Q = 5 \times 10^{- 5}$
$∴ Q$ on $4 μ F$ is $5 \times 10^{- 5}$
$∴ V = \frac{Q}{C}$
$V = \frac{5 \times 10^{- 5}}{4 \times 10^{- 6}}$
$V = 12.5 V$

Two condensers of capacitance 4μF and 5μF are joined in series. If the potential difference across 5μF is 10V, then the potential difference across 4μF condenser is :22.5V10V12.5V25V

When two capacitors are in series Q present on then is same∴ Q on 5μF is Q=5×10−6×10Q=5×10−5∴Q on 4μF is 5×10−5∴V=QCV=5×10−54×10−6V=12.5 V

Two condensers of capacitance $4 μ F$ and $5 μ F$ are joined in series. If the potential difference across $5 μ F$ is $10 V$ , then the potential difference across $4 μ F$ condenser is :

$22.5 V$
$10 V$
$12.5 V$
$25 V$

When two capacitors are in series $Q$ present on then is same
$∴ Q o n 5 μ F$ is $Q = 5 \times 10^{- 6} \times 10$
$Q = 5 \times 10^{- 5}$
$∴ Q$ on $4 μ F$ is $5 \times 10^{- 5}$
$∴ V = \frac{Q}{C}$
$V = \frac{5 \times 10^{- 5}}{4 \times 10^{- 6}}$
$V = 12.5 V$