Two capacitors of capacities c1=3uf and c2=6uf arranged in series are connected in a parallel with third capacitor c3=4uf. The arrangement is connected to a 6.0 V battery. Calculate the total energy stored in the capacitors.
C = C1*C2/(C1+C2) = 3*6/(3+6) = 2 uF.
Ct = C + C3 = 2 + 4 = 6 uF.
Total Energy = 0.5C*E^2 = 0.5*6*6^2 = 108 J.
In the numerical value, C is value is 6 uf, otherwise it's OK
All was correct except si unit to micro joule
To calculate the total energy stored in the capacitors, we first need to find the equivalent capacitance of the series and parallel combination of capacitors.
1. Series Combination:
When capacitors are connected in series, the reciprocal of the equivalent capacitance is given by the sum of the reciprocals of the individual capacitances.
So, the equivalent capacitance of the series combination (Cs) is given by:
1/Cs = 1/c1 + 1/c2
Substituting the given values of c1 and c2:
1/Cs = 1/3µF + 1/6µF
1/Cs = (2 + 1)/6µF
1/Cs = 1/2µF
Therefore, Cs = 2µF
2. Parallel Combination:
When capacitors are connected in parallel, the equivalent capacitance is simply the sum of the individual capacitances.
So, the equivalent capacitance of the parallel combination (Cp) is given by:
Cp = c3
Substituting the given value of c3:
Cp = 4µF
3. Total Energy Stored:
The energy stored in a capacitor is given by the formula:
E = (1/2) * C * V^2
The energy stored in the series combination (Es) is given by:
Es = (1/2) * Cs * V^2
Substituting the given value of V (6.0V) and Cs (2µF):
Es = (1/2) * 2µF * (6.0V)^2
Es = (1/2) * 2µF * (36V^2)
Es = 72µJ
The energy stored in the parallel combination (Ep) is given by:
Ep = (1/2) * Cp * V^2
Substituting the given value of V (6.0V) and Cp (4µF):
Ep = (1/2) * 4µF * (6.0V)^2
Ep = (1/2) * 4µF * (36V^2)
Ep = 432µJ
The total energy stored in the capacitors (Et) is the sum of the energies stored in the series and parallel combinations:
Et = Es + Ep
Et = 72µJ + 432µJ
Et = 504µJ
Therefore, the total energy stored in the capacitors is 504µJ.