Two capacitors of capacitance C1= 1.5 x 10-6 F and C2 = .5 x 10-6 F are connecte

Question

Two capacitors of capacitance C1= 1.5 x 10-6 F and C2 = .5 x 10-6 F are connected to an 8 V battery as shown in the figure below. Initially both capacitors are uncharged. When the switch S is turned to position 1, the plates of the capacitor C1 acquire a potential difference.The switch is then turned to position 2.

What are the final charges q1 and q2 on the corresponding capacitors?

What is the potential difference across C2?

Two capacitors of capacitance C1= 1.5 x 10-6 F and C2 = .5 x 10-6 F are connected to an 8 V battery as shown in the figure below. Initially both capacitors are uncharged. When the switch S is turned to position 1, the plates of the capacitor C1 acquire a potential difference.The switch is then turned to position 2. What are the final charges q1 and q2 on the corresponding capacitors? What is the potential difference across C2?

Explanation / Answer

when switch is turn to position 1 then only C1 is connected to voltage source so total charge on capacitor C1 is given by

C=Q/V
Q=C*V

Q=8*1.5*10^-6

Q1=12*10^-6 C

when switch is turned to position 2 the capacitor C1 and C2 are in the circuit without any source only C1 capacitor charge so the capacitor C1 discharge untill when the charge on C2 and C1 are equal

so Charge on C2 capacitor is (12*10^-6)/2

Q2= 6*10^-6 C

therefore the voltage across it given as

C= Q/V

V=Q/C

V2= Q2/C2

V2=6*10^-6/(.5*10^-6)

V2=12 V

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