Question

Two capacitors of capacitance 4.0 F and 6.0 F are charged to a potential difference of 12.0 V and 6.0 V, respectively. If the capacitors were arranged in a parallel connection, calculate the total energy stored.

Answer 1

Given that,

Capacitance of one capacitor = 4.0 F

Capacitance of another capacitor = 6.0 F

Potential difference V₁= 12.0 V

Potential difference V₂ = 6.0 V

We need to calculate the charge on one capacitor

Using formula of charge

$Q_{1}=C_{1}V_{1}$

Put the value into the formula

$Q_{1}=4.0\times12.0$

$Q_{1}=48.0\ C$

We need to calculate the charge on another capacitor

Using formula of charge

$Q_{2}=C_{2}V_{2}$

Put the value into the formula

$Q_{2}=6.0\times6.0$

$Q_{2}=36.0\ C$

We need to calculate the common potential

Using formula of potential

$V=\dfrac{total\ charge}{total\ capacitors}$

$V=\dfrac{Q_{1}+Q_{2}}{C_{1}+C_{2}}$

Put the value into the formula

$V=\dfrac{48+36}{4+6}$

$V=\dfrac{42}{5}$

We need to calculate the total energy stored

Using formula of energy

$E=\dfrac{1}{2}\times(C_{1}+C_{2})V^2$

Put the value into the formula

$E=\dfrac{1}{2}\times(4+6)\times(\dfrac{42}{5})^2$

$E=352.8\ J$

Hence, The total energy stored is 352.8 J.