a capacitance of 6microfarad and another unknown capacitance are connected in series to a battery, if the potential difference between the battery terminals is 16v while over the unknown capacitance is 6v, find the magnitude of the unknown capacitance, the total energy furnished by the battery and the total energy in the charged capacitance
C = Q/V so V = Q/C
Vk = 10
Vu = 6
Q the same on both
so
10 = Q/6*10*-6
Q = 6*10^-5 coulombs
6 = 6*10^-5 /Cu
Cu = 10^-5 farads = 10 microfarad
energy = (1/2) Q V for each
What about the third one that is the total energy furnished
To find the magnitude of the unknown capacitance, we can use the formula for capacitors in series:
1/C_total = 1/C_1 + 1/C_2
Let C_2 be the unknown capacitance.
1/C_total = 1/6μF + 1/C_2
Since we know that the potential difference over the unknown capacitance is 6V, we can use the formula for the energy stored in a capacitor:
E = (1/2) * C * V^2
First, let's find the total capacitance (C_total):
1/C_total = 1/6μF + 1/C_2
To combine the fractions, we need to find a common denominator:
1/C_total = (C_2 + 6μF) / (6μF * C_2)
To find the reciprocal:
C_total = (6μF * C_2) / (C_2 + 6μF)
Now, we can substitute the values and solve for C_2:
C_total = (6μF * C_2) / (C_2 + 6μF)
16μF = (6μF * C_2) / (C_2 + 6μF)
Now, let's solve this equation for C_2. Multiply both sides by (C_2 + 6μF):
16μF * (C_2 + 6μF) = 6μF * C_2
Expand the equation:
16μF * C_2 + 96μF^2 = 6μF * C_2
Simplify the equation:
10μF * C_2 = 96μF^2
Divide both sides by 10μF:
C_2 = 9.6μF
So, the magnitude of the unknown capacitance is 9.6μF.
To find the total energy furnished by the battery, we can use the formula:
E_total = (1/2) * C_total * V^2
E_total = (1/2) * 16μF * (16V)^2
E_total = 128μJ
The total energy furnished by the battery is 128μJ.
To find the total energy in the charged capacitance, we can use the formula:
E_C2 = (1/2) * C_2 * V^2
E_C2 = (1/2) * 9.6μF * (6V)^2
E_C2 = 172.8μJ
The total energy in the charged capacitance is 172.8μJ.
To solve this problem, we can derive the equations governing capacitors in series.
Let's start by calculating the unknown capacitance. In a series circuit, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of the individual capacitances:
1/C_total = 1/C_1 + 1/C_2
where C_1 represents the known capacitance (6µF) and C_2 represents the unknown capacitance.
Rearranging the equation, we have:
1/C_2 = 1/C_total - 1/C_1
Substituting the known values:
1/C_2 = 1/C_total - 1/6µF
Since the potential difference over the unknown capacitance is 6V, we can relate the charge (Q) on the unknown capacitance and the potential difference (V) using the formula:
Q = C_2 * V
We know that the potential difference is 6V. Therefore, we have:
Q = C_2 * 6V
Next, we can calculate the total energy furnished by the battery. The energy stored in a capacitor is given by the formula:
E = 0.5 * C * V^2
where E represents the energy, C is the capacitance, and V is the potential difference.
Substituting the known values, the energy stored in the unknown capacitance becomes:
E_2 = 0.5 * C_2 * (6V)^2
Finally, the total energy furnished by the battery is the sum of the energy stored in the known and unknown capacitors:
Total Energy = Energy in C_1 + Energy in C_2
Now, you can plug in the given values and calculate the desired quantities.