Two capacitors, one 12.0 𝜇𝐹 and the other of unknown capacitance C, are connected in parallel across a battery with an emf of 9.00 V. The total energy stored in the two capacitors is 0.0115 J. Determine the capacitance C.a.272 𝜇𝐹b.180 𝜇𝐹c.12 𝜇𝐹d.284𝜇𝐹
Question
Two capacitors, one 12.0 𝜇𝐹 and the other of unknown capacitance C, are connected in parallel across a battery with an emf of 9.00 V. The total energy stored in the two capacitors is 0.0115 J. Determine the capacitance C.a.272 𝜇𝐹b.180 𝜇𝐹c.12 𝜇𝐹d.284𝜇𝐹
Solution
The energy stored in a capacitor can be calculated using the formula:
E = 1/2 * C * V^2
where E is the energy, C is the capacitance, and V is the voltage.
Given that the total energy stored in the two capacitors is 0.0115 J and the voltage across the capacitors is 9.00 V, we can set up the following equation:
0.0115 J = 1/2 * 12.0 μF * (9.00 V)^2 + 1/2 * C * (9.00 V)^2
Solving for C gives:
C = (0.0115 J - 1/2 * 12.0 μF * (9.00 V)^2) / (1/2 * (9.00 V)^2)
Calculating the above expression should give the value of the unknown capacitance C.
Let's calculate it:
First, calculate the energy stored in the 12.0 μF capacitor:
E1 = 1/2 * 12.0 μF * (9.00 V)^2 = 0.00486 J
Subtract this from the total energy to find the energy stored in the capacitor with unknown capacitance:
E2 = 0.0115 J - 0.00486 J = 0.00664 J
Now, use this energy to find the unknown capacitance:
C = 2*E2 / (9.00 V)^2 = 0.0001644 F = 164.4 μF
So, the unknown capacitance is approximately 164.4 μF, which is not listed in the given options. There might be a mistake in the problem or the given options.
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