A series arrangement of three capacitor of value 8 microfarad 12 microfarad and 24 microfarad is connected in series with a 90 volt battery (1) draw an open circuit diagram for this arrangement (2) calculate the effective capacitance in the circuit (3) on close circuit calculate the charge on each capacitor when fully charged (4) determine the potential difference across the 8 microfarad capacitor
1)
Open Circuit Diagram:
90V Battery
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C1 (8 μF) -- C2 (12 μF) -- C3 (24 μF)
2) Effective capacitance in the circuit = 8 μF + 12 μF + 24 μF = 44 μF
3) Charge on each capacitor when fully charged:
C1 (8 μF) = 8 μF x 90V = 720 μC
C2 (12 μF) = 12 μF x 90V = 1080 μC
C3 (24 μF) = 24 μF x 90V = 2160 μC
4) Potential difference across the 8 μF capacitor = 8 μF x 90V = 720 V
In the electrolysis of one of it's salt 0.594 g of a metal were deposited by a current of 0.5A flowing for 60 minutes. Calculate the mass of metal which would be deposited by a current of 0.4 A flowing for 40 minutes.
1/C = 1/C1 + 1/C2 + 1/C3
when in series
The CHARGE is the same on each (otherwise where would it come from)
(1) To draw an open circuit diagram for this arrangement, you would simply place the three capacitors in series with each other and connect them in series with the 90V battery. The positive terminal of the battery will be connected to the positive terminal of the first capacitor, and the negative terminal of the battery will be connected to the negative terminal of the last capacitor. The diagram should look like this:
Battery (+90V) - Capacitor (8µF) - Capacitor (12µF) - Capacitor (24µF) - Battery (-90V)
(2) To calculate the effective capacitance in the circuit, you would use the formula for the equivalent capacitance of capacitors in series:
1/Ceq = 1/C1 + 1/C2 + 1/C3
Where C1, C2, and C3 are the capacitances of the three capacitors.
Substituting the given values, we have:
1/Ceq = 1/8µF + 1/12µF + 1/24µF
Simplifying this expression gives us the reciprocal of the effective capacitance. Taking the reciprocal of both sides, we find:
Ceq = 1 / (1/8µF + 1/12µF + 1/24µF)
Calculating further will give us the value of the effective capacitance.
(3) When the circuit is closed and the capacitors are charging, the total charge in the circuit will be the same as the charge on each capacitor. To calculate the charge on each capacitor when fully charged, we can use the formula:
Q = Ceq * V
Where Q is the charge, Ceq is the effective capacitance, and V is the voltage across the capacitors (which is equal to the battery voltage, 90V in this case). By substituting the known values into the formula, you can calculate the charge on each capacitor.
(4) To determine the potential difference across the 8µF capacitor, you can use the formula for the voltage across a capacitor in a series circuit:
Vc1 = V * (C1 / Ceq)
Where Vc1 is the potential difference across the 8µF capacitor, V is the battery voltage (90V), C1 is the capacitance of the 8µF capacitor, and Ceq is the effective capacitance of the circuit (calculated in question 2).
By substituting the known values into the formula, you will be able to find the potential difference across the 8µF capacitor.