Question

A 200.0V battery is connected across two capacitor of capacitance 4.0 micro fried and 6.0 micro fried in parallel calculate the energy store in each capacitor and energy supplied by the store.​

Answer 1

• Capacitance, C_1 = 4 μF

• Capacitance, C_2 = 6 μF

• Potential difference of the battery, V = 200 v

Since the capacitors are connected in parallel, the potential difference is same for both capacitors.

Energy stored in the capacitor, E = CV²/2

Consider for 4 μF capacitor

$E = \frac{1}{2} C_1 {V}^{2} \\ \\ = \frac{1}{2} \times 4 \times {(200)}^{2} \\ \\ = \frac{1}{2} \times 4 \times 200 \times 200 \\ \\ = 4 \times 200 \times 100 \\ \\ E = 80000 \: \mu F \\ \\ E = 80 \times 1000 \times {10}^{ - 6} \: F \\ \\ = 80 \times {10}^{3} \times {10}^{ - 6} \\ \\ = 80 \times {10}^{(3 - 6)} \\ \\ = 80 \times {10}^{ - 3} \\ \\ E = 80 \: mF$

Energy stored in the 4 μF capacitor is 80 mF.

Consider for 6 μF capacitor

$E = \frac{1}{2} C_2 {V}^{2} \\ \\ = \frac{1}{2} \times 6 \times {(200)}^{2} \\ \\ = \frac{1}{2} \times 6 \times 200 \times 200 \\ \\ = 6 \times 200 \times 100 \\ \\ E = 120000 \: \mu F \\ \\ E = 120 \times 1000 \times {10}^{ - 6} \: F \\ \\ = 120 \times {10}^{3} \times {10}^{ - 6} \\ \\ = 120 \times {10}^{(3 - 6)} \\ \\ = 120 \times {10}^{ - 3} \\ \\ E = 120 \: mF$

Energy stored in the 6 μF capacitor is 120 mF.

Total energy stored is equal o the sum of the individual energies of capacitors.

E = 80 + 120

E = 200 mF

ALTERNATIVE METHOD FOR TOTAL ENERGY:

Since the capacitors are connected in series, the effective capacitance is equal to the sum of individual capacitances.

$C_{eff} = C_1 + C_2 \\ \\ = 4 + 6 \\ \\ C_{eff} = 10 \: \mu F$

The net capacitance is 10 μF.

Total energy stored,

$E = \frac{1}{2} \times C_{eff} \times {V}^{2} \\ \\ = \frac{1}{2} \times 10 \times {(200)}^{2} \\ \\ = \frac{1}{2} \times 10 \times 200 \times 200 \\ \\ = 10 \times 200 \times 100 \\ \\ E = 200000 \: \mu F \\ \\ = 200 \times 1000 \times {10}^{ - 6 } F \\ \\ = 200 \times {10}^{3} \times {10}^{ - 6} \\ \\ = 200 \times {10}^{ - 3} \\ \\ E = 200 \: mF$

The total energy stored is 200 mF.