Question

Two capacitors are connected to a battery. The battery voltage is V = 90 V, and the capacitances are C1 = 2.00 μF and C2 = 4.00 μF. Determine the total energy stored by the two capacitors when they are wired (a) in parallel and (b) in series.

Answer 1

Answer:
`E=24.3* 10^(-3) J`

`E=5.4* 10^(-3) J`

Step-by-step explanation:
`E=24.3* 10^(-3) J`

`E=5.4* 10^(-3) J`

Given

`C_1=2 \mu F`

`C_2=4\mu F`

Voltage
`\left ( V\right )=90 V`

`\left ( a\right ) parallel`

`C_(eq)=C_1+C_2=2+4=6\mu F`

Energy stored in capacitor
`\left ( E\right )=(1)/(2)C_(eq)V^2`

`E=(1)/(2)* 6* 10^(-6)* 90^2`

`E=24.3* 10^(-3) J`

`\left ( b\right )Series`

`C_(eq)=(C_1C_2)/(C_1+C_2)`

`C_(eq)=(4)/(3) \mu F`

Energy stored=
`(1)/(2)C_(eq)V^2`

E=
`(1)/(2)* (4)/(3)* \left ( 90\right )^2`

E=
`5.4* 10^(-3) J`