Could someone check my Electronics answers?

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SUMMARY

The forum discussion focuses on solving two electronics problems related to capacitors and circuits. In Question 1, the user calculates the resistance of a circuit and the potential difference across a capacitor, concluding that the resistance is 0.1 Ohms and the potential difference is 10 Volts. In Question 2, the user determines that a 10 microfarad capacitor charged by a 24V battery stores 2.4 x 10^-4 Coulombs of charge and 2.88 x 10^-3 Joules of energy. The user seeks clarification on the energy required from the battery compared to the energy stored in the capacitor.

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I've got my final Electronics paper to hand in on Thursday and would be grateful is someone could give my answers the once over :)


Question 1.

The circuit (the attached picture) is connected up with switch S closed. The variable resistor, R is adjusted so that the micro-ammeter reads 100mf (micro farads). For this question you should assume that the micro-ammeter has negligible resistance.

A) With switch S closed,
i) Calculate the resistor of the circuit connected to the battery
ii) State the Potential Difference across the capacitor

B) When switch S is opened and the circuit is not adjusted, state briefly what happens.

C) After switch S is opened the variable resistor, R is gradually adjusted to maintain a constant current 100mf (micro farads). The current is sustained for 10s. After 10s no more current flows.
i) Calculate the charge stored in the capacitor at the moment the switch was opened
ii) Calculate the capacitance of the capacitor.

Question 2. A 10mf (micro farads) capacitor is fully charged by connecting it to a 24 V battery

i) How much charge would be stored in the capacitor?
ii) How much energy would be stored in the capacitor?
iii) Hoe much energy would be required from the battery to charge the capacitor to the 24v level?
iv) Explain why your answer to part (iii) is different from part (ii)



Answer Question 1
A)
i) V = I*R
R = V / I
R = 10V / 100mf
R = 0.1 Ohms

ii) 10Volts

B) When the switch is opened, the circuit is broken and the capacitor begins to drain.


C)
i) Q = DV/Dt
Q = 10V / 10s
Q = Q
Q = 1C

ii) Q/V = C
1/10F = C

Question 2.

A)
i) Q = V *C
Q = 24*10*10^-6
Q = 2.4 * 10^-4 C

ii) E = 1/2 Q*V

E = 1/2 * 2.4x10^-4 *24
E = 2.88 x 10^-3 J

iii) E = 2.88 x 10^-3 J / 24x10^-4 C
E = 12J

iv) ...Really not sure about this! I know its basic so could anyone give me a hint? :)

Thanks - Really appreciate it!
 

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...Would really appreciate it :)
 
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