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)
Part B: Calculate the charge on each capacitor.
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
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
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
Part A: If V = 16V is applied across the whole network(Figure 1) , calculate the...
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