Question

The figure shows a network of fuve capacitors connected to a 10v battery. Calculate the charge acquired by the 5uF capscitor.

The figure shows a network of fuve capacitors connected to a 10v battery. Calculate the charge acquired by the 5uF capscitor. 20 10 V The figure shows a network of five capacitors connected Calculate the charge acquired by the 5 PF capacitor. 5 10 15 10 20 II 10 v, m *tr$ra 10 67 1 I o•ls

Answer 1

Answer:

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Answer 2

Given that.

Voltage = 10 V

Capacitor $C_{1}=15\ \mu F$

Capacitor $C_{2}=5\ \mu F$

Capacitor $C_{3}=10\ \mu F$

Capacitor $C_{4}=10\ \mu F$

Capacitor $C_{5}=20\ \mu F$

Voltage across AC and A'C' is same.

We need to calculate the equivalent capacitance between A and B

Using formula of parallel

$C_{AB}=C_{1}+C_{2}$

Put the value into the formula

$C_{AB}=15+5$

$C_{AB}=20\ \mu F$

We need to calculate the equivalent capacitance between B and C

Using formula of parallel

$C_{BC}=C_{3}+C_{4}$

Put the value into the formula

$C_{BC}=10+10$

$C_{BC}=20\ \mu F$

We need to calculate the equivalent capacitance between A and C

Using formula of series

$\dfrac{1}{C_{AC}}=\dfrac{1}{C_{AB}}+\dfrac{1}{C_{BC}}$

Put the value into the formula

$\dfrac{1}{C_{AC}}=\dfrac{}{20}+\dfrac{1}{20}$

$C_{AC}=10\ \mu F$

We need to calculate the charge in $C_{AC}$

Using formula of charge

$Q=CV$

Where, C = capacitors

V = voltage

Put the value into the formula

$Q=10\times10$

$Q=100\ \mu C$

We need to calculate the voltage drop across AB

Using formula of voltage

$V_{AB}=\dfrac{Q}{C_{AB}}$

Put the value into the formula

$V_{AB}=\dfrac{100}{20}$

$V_{AB}=5\ V$

We need to calculate the charge acquired by the 5 μF capacitor

Using formula of charge

$Q=CV$

Put the value into the formula

$Q=5\times5$

$Q=25\ \mu C$

Hence,  The charge acquired by the 5 μF capacitor is 25 μC.