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Physics Ncert Solutions Class 12th Physics Class 12th Electrostatic Potential and Capacitance

2.25 Obtain the equivalent capacitance of the network in Fig. 2.33. For a 300 V supply, determine the charge and voltage across each capacitor.

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Answered by

alok kumar singh | Contributor-Level 10

5 months ago

2.25 Let C' be the equivalent capacitance for capacitors $C_{2}$ and $C_{3}$ connected in series.

Hence, $\frac{1}{C'}$ = $\frac{1}{200}$ + $\frac{1}{200}$ . So C' = 100 pF

Capacitors C' and $C_{1}$ are in parallel, if the equivalent capacitance be C”, then

C” = C' + $C_{1}$ = 100 + 100 = 200 pF

Now C” and $C_{4}$ are connected in series. If the total equivalent capacitance of the circuit be $C_{T o t a l}$ , then $\frac{1}{C_{T o t a l}}$ = $\frac{1}{C "}$ + $\frac{1}{C_{4}}$ = $\frac{1}{200}$ + $\frac{1}{100}$ , $C_{T o t a l}$ = $\frac{200}{3}$ pF = $\frac{200}{3} * 10^{- 12}$ F

Let V” be the potential difference across C” and $V_{4}$ be the potential difference across $C_{4}$

Then, V” +

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2.25 Let C' be the equivalent capacitance for capacitors $C_{2}$ and $C_{3}$ connected in series.

Hence, $\frac{1}{C'}$ = $\frac{1}{200}$ + $\frac{1}{200}$ . So C' = 100 pF

Capacitors C' and $C_{1}$ are in parallel, if the equivalent capacitance be C”, then

C” = C' + $C_{1}$ = 100 + 100 = 200 pF

Let V” be the potential difference across C” and $V_{4}$ be the potential difference across $C_{4}$

Then, V” + $V_{4}$ = 300 V

Now, charge $Q_{4}$ on $C_{4}$ is given by $Q_{4}$ = $C_{T o t a l} * V$ = $\frac{200}{3} * 10^{- 12}$ $* 300$ = 2 $* 10^{- 8}$ C

$V_{4} = \frac{Q_{4}}{C_{4}}$ = $\frac{2 * 10^{- 8}}{100 * 10^{- 12}}$ = 200. So V” = 100 V

So potential difference across C' and $C_{1}$ is V” = 100 V

Charge on $C_{1}$ is given by $Q_{1}$ = 100 $* 10^{- 12} * 100$ C = 1 $* 10^{- 8}$ C

$C_{2}$ and $C_{3}$ are having same capacitance and have a potential difference of 100V together. Since $C_{2}$ and $C_{3}$ are in series, the potential difference is given by $V_{2}$ = $V_{3}$ = 50V

Hence the charge on $C_{2}$ is given by $Q_{2}$ = 200 $* 10^{- 12} * 50$ = $10^{- 8}$ C and the charge on $C_{3}$ is given by $Q_{3}$ = 200 $* 10^{- 12} * 50$ = $10^{- 8}$ C

Therefore, the equivalent capacitance of the circuit is $\frac{200}{3}$ pF and charge and voltage at all capacitance is given as

For $C_{1 :} Q_{1}$ = $10^{- 8}$ C, $V_{1}$ = 100 V

For $C_{2 :} Q_{2}$ = $10^{- 8}$ C, $V_{2}$ = 50 V

For $C_{3 :} Q_{3}$ = $10^{- 8}$ C, $V_{3}$ = 50 V

For $C_{4 :} Q_{4}$ = 2 $* 10^{- 8}$ C, $V_{4}$ = 200 V

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Answered by

alok kumar singh | Contributor-Level 10

3 months ago

2.25 Let C’ be the equivalent capacitance for capacitors $C_{2}$ and $C_{3}$ connected in series.

Hence, $\frac{1}{C'}$ = $\frac{1}{200}$ + $\frac{1}{200}$ . So C’ = 100 pF

Capacitors C’ and $C_{1}$ are in parallel, if the equivalent capacitance be C”, then

C” = C’ + $C_{1}$ = 100 + 100 = 200 pF

Let V” be the potential difference across C” and $V_{4}$ be the

...more

2.25 Let C’ be the equivalent capacitance for capacitors $C_{2}$ and $C_{3}$ connected in series.

Hence, $\frac{1}{C'}$ = $\frac{1}{200}$ + $\frac{1}{200}$ . So C’ = 100 pF

Capacitors C’ and $C_{1}$ are in parallel, if the equivalent capacitance be C”, then

C” = C’ + $C_{1}$ = 100 + 100 = 200 pF

Let V” be the potential difference across C” and $V_{4}$ be the potential difference across $C_{4}$

Then, V” + $V_{4}$ = 300 V

Now, charge $Q_{4}$ on $C_{4}$ is given by $Q_{4}$ = $C_{T o t a l} \times V$ = $\frac{200}{3} \times 10^{- 12}$ $\times 300$ = 2 $\times 10^{- 8}$ C

$V_{4} = \frac{Q_{4}}{C_{4}}$ = $\frac{2 \times 10^{- 8}}{100 \times 10^{- 12}}$ = 200. So V” = 100 V

So potential difference across C’ and $C_{1}$ is V” = 100 V

Charge on $C_{1}$ is given by $Q_{1}$ = 100 $\times 10^{- 12} \times 100$ C = 1 $\times 10^{- 8}$ C

$C_{2}$ and $C_{3}$ are having same capacitance and have a potential difference of 100V together. Since $C_{2}$ and $C_{3}$ are in series, the potential difference is given by $V_{2}$ = $V_{3}$ = 50V

Hence the charge on $C_{2}$ is given by $Q_{2}$ = 200 $\times 10^{- 12} \times 50$ = $10^{- 8}$ C and the charge on $C_{3}$ is given by $Q_{3}$ = 200 $\times 10^{- 12} \times 50$ = $10^{- 8}$ C

Therefore, the equivalent capacitance of the circuit is $\frac{200}{3}$ pF and charge and voltage at all capacitance is given as

For $C_{1 :} Q_{1}$ = $10^{- 8}$ C, $V_{1}$ = 100 V

For $C_{2 :} Q_{2}$ = $10^{- 8}$ C, $V_{2}$ = 50 V

For $C_{3 :} Q_{3}$ = $10^{- 8}$ C, $V_{3}$ = 50 V

For $C_{4 :} Q_{4}$ = 2 $\times 10^{- 8}$ C, $V_{4}$ = 200 V

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2.25 Obtain the equivalent capacitance of the network in Fig. 2.33. For a 300 V supply, determine the charge and voltage across each capacitor.

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