To solve this problem, let's first consider the initial state of the circuit, where switch A is closed and switch B is open. In this case, the 3uF capacitor is fully charged and the 5uF capacitor is uncharged (Q = 0).
When switch A is opened and switch B is closed, the circuit configuration changes. The two capacitors are now connected in series, which means that the total capacitance of the circuit is given by the reciprocal of the sum of the reciprocals of the individual capacitances:
C_total = 1 / (1/C1 + 1/C2)
= 1 / (1/3 + 1/5)
= 1 / (5/15 + 3/15)
= 1 / (8/15)
= 15/8 uF
= 1.875 uF
Now, let's calculate the final potential difference across the combination when switch B is closed. Since the capacitors are in series, the charge stored in each capacitor is the same. Let's assume this charge is Q.
Q = C_total * V_final
Since the net charge must be conserved, we can write:
Q = C1 * V1 = C2 * V2
Since the 5uF capacitor was initially uncharged, we have V2 = 0. Let's substitute this into the equation:
Q = C1 * V1 = C2 * 0 = 0
Now, let's solve for Q:
Q = C1 * V1
Q = (3uF) * V1
Substituting back into the equation:
(3uF) * V1 = 0
Therefore, the potential difference across the combination when switch B is closed is 0V.
Now, let's calculate the final energy stored by the capacitors. The energy stored by a capacitor is given by the formula:
E = (1/2) * C * V^2
The initial energy stored by the 3uF capacitor is given as 5.4 * 10^5 J. Let's use this information to find the final energy.
Initial energy stored by 3uF capacitor = (1/2) * C1 * V1^2
Final energy stored by 3uF capacitor = (1/2) * C1 * 0^2 = 0 J
The final energy stored by the 5uF capacitor is also 0 J, since it was uncharged initially.
Therefore, the final energy stored by the capacitors is 0 J.
The difference in the initial and final values of the energy stored is due to the fact that the final potential difference across the combination is 0V, which means that no charge is stored in the capacitors and hence no energy is stored.
A circuit where switch A is initially closed and switch B is open. When the 3uF capacitor is fully charged, switch A is opened and switch B is closed.connected to a battery of emf 6V. Another capacitor of 5uF is connected in series with the 3uF capacitor.calculate the final potential difference across the combination when switch B is closed and final energy storage by the capacitors
Account for the difference in the initial and final values of the energy stored , given that the intial energy stored by the 3uF capacitor when fully charged is 5,4×10^5J.
1 answer
1 answer