High School

If [tex]C_1 = C_2 = 4.00 \mu F[/tex] and [tex]C_4 = 8.00 \mu F[/tex], what should the capacitance [tex]C_3[/tex] be for the network to store [tex]2.70 \times 10^{-3} \, J[/tex] of electrical energy?

Answer :

The value of the third capacitance is determined as 8 μF.

How calculate the third capacitance?

The value of the third capacitance is calculated by applying the following formula as shown below.

The formula for energy stored in a capacitor;

E = ¹/₂ CV²

where;

  • C is the equivalent capacitance of the four capacitors
  • V is the voltage in the circuit

The equivalent capacitance of the four capacitors in parallel is calculated as;

CV² = 2E

C = 2E/V²

C = ( 2 x 2.7 x 10⁻³ ) / ( 15²)

C = 2.4 x 10⁻⁵ F

C = 24 x 10⁻⁶ F

C = 24 μF

The value of the third capacitance is calculated as;

C3 = C - (C1 + C2 + C4)

C3 = 24 μF - (4μF + 4μF + 8μF)

C3 = 8 μF

The complete question is below:

Four capacitors are in parallel, C1, C2, C3, and C4. If C1​=C2​=4.00μF and C4​=8.00μF, what should the capacitance C3​ be for the network to store 2.70×10⁻³J of electrical energy at a 15 Volts?