Question

Part A: If V = 16V is applied across the whole network(Figure 1) , calculate the voltage across each capacitor. Express your answers using three significant figures separated by commas. (**Answer should be: v1, v2, v3)

If V = 16V is applied across the whole network(Figure 1) , calculate the voltage across each capacitor. Express your answers using three significant figures separated by commas. (**Answer should be: v1, v2, v3)

Part B: Calculate the charge on each capacitor.

Express your answers using three significant figures separated by commas.

Answer 1

voltage across C1 = V*4/(3+4)

= 16*4/7

=9.14 V

voltage across C2 = 16 -9.14

= 6.86 V

voltage across C3 = V

= 16V

answer is 9.14 , 6.86 , 16.0

PART B

Q3 = C3V3

= 2*16 uC

= 32 uC

Q2= C2*V2

= 3*6.86 uC

= 20.58 uC

Q1= C1*V1

= 4* 9.14 uC

=36.56 uC

charge is 35.6 uC , 20.6 uC , 32.0 uC

Answer 2

part A:

C1 and C2 are in Series

and C1 and C2 combined is in parallel with C3

equivalent capacitance between C1 and C2=(C1*C2)/(C1+C2)=12/7 uF

So the voltage across C3=16V

and voltage acoss C1 and C2 combined =16V

C1V1=C2V2

and V1+V2=16

solving we get V1=64/7 V

V2=48/7 V

Part B:

Charge on C1=Charge on C2=C1V1=C2V2=192/7 C=27.428 uC

Charge on C3=C3*V3=2*16=32 uC

Answer 3

Let the equivalent capacitance across C1 and C2 be C

so... 1/C = 1/C1 + 1/C2

so.. 1/C = 1/3 + 1/4

so.. C = 1.714285714286 uC

equivalent capacitance across C and C3 = C + C3 = 1.714285714286 + 2 = 3.714285714286 uC

so.. voltage across C = 16 * C3 / ( C + C3 ) = 16 * 2 / (3.714285714286) = 8.615384615384 V

so.. voltage across C1 = voltage across C2 = v1 = v2 = 8.615384615384 V

voltage actross C3 = v3 = 16 - 8.615384615384 = 7.384615384616 V

part B

charge on C1 = q1 = C1 * v1 = 3 * 10^-6 * 8.615384615384 = 25.846153846152 uC

charge on C2 = q2 = C2 * v2 = 4 * 10^-6 * 8.615384615384 = 34.461538461536 uC

charge on C1 = q1 = C1 * v1 = 2 * 10^-6 * 7.384615384616 = 14.769230769232 uC