Two capacitors C1 (6uF) and C2 (12uF) are in series across a 180 volts d.c. supply. Calculate the charges on C1 and C2 respectively.
Correct answer: D. 72 x 10^-5 C, 72 x 10^-5 C
- A. 120 x 10^-6 C, 420 x 10^-6 C
- B. 320 x 10^-6 C, 420 x 10^-6 C
- C. 420 x 10^-6 C, 320 x 10^-6 C
- D. 72 x 10^-5 C, 72 x 10^-5 C
- E. 820 x 10^-6 C, 420 x 10^-6 C
Explanation
The formula for calculating the charge on a capacitor is:Charge = Capacitance * VoltageIn this case, the voltage across both capacitors is the same, so the charge on both capacitors is also the same. However, the capacitance of C1 is half that of C2. This means that the voltage across C1 must be twice the voltage across C2.To calculate the charge on each capacitor, we can use the following equations:Charge_C1 = 6e-6 * 2 * 180 = 72 x 10-5CCharge_C2 = 12e-6 * 180 = 72 x 10-5CAs you can see, the charges on both capacitors are the same, which is 72 x 10-5CThe reason why the voltage across C1 is twice the voltage across C2 is because the capacitance of C1 is half that of C2. The voltage across a capacitor is inversely proportional to its capacitance. This means that if the capacitance of a capacitor is halved, the voltage across the capacitor will double.
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