T-Mobile LTE 3:29 PM Four capacitors are connected as shown in the figure below.

Question

T-Mobile LTE 3:29 PM Four capacitors are connected as shown in the figure below. (Let C = 20.0 HF.) 6.00 ph (a) Find the equivalent capacitance between points a and b. HF (b) Calculate the charge on each capacitor, taking AVab 18.0 V 3.00 capacitor uC Find the following. (In the figure, use C11.40 F and C-5.40 F) 6.00p (a) the equivalent capacitance of the capacitors in the figure above LF (b) the charge on each capacitor en the right 11.0-F capacter on the left 1140-gif capactorC on the 5.40- capacar on the 6.00 cpacit (c) the potential difference across each capacitor on the right 3.40-F capacitor en the left 11.0-capac on the 6.00-cpc Need Help? ?hnu

Explanation / Answer

12.

Remember:
For parallel combination
Ceq = C1 + C2 + C3 +...............
for series combination
1/Ceq = 1/C1 + 1/C2 + 1/C3 + ............
for 2 capacitors in series it will be
Ceq = C1*C2/(C1+C2)
Using this Information:

C2 and 6 uF are in parallel, So

Cp = C2 + 6 = 5.40 + 6.00 = 11.40 uF

Now Cp, C1 and C1 are in series, So

Ceq = (1/Cp + 1/C1 + 1/C1)^-1

Ceq = (1/11.40 + 1/11.40 + 1/11.40)^-1

Ceq = 11.40/3 = 3.80 uF

Now we know that

Qeq = Ceq*V

Qeq = 3.80*9.00 = 34.2 uC

Now remember in capacitors parallel combination voltage distribution in each part will be same and in series combination charge distribution in each capacitor will be same.

Charge on C1 (right) = Charge on C1 (left) = charge on Cp = 34.2 uC

Charge on C1 (right) = 34.2 uC

Charge on C1 (left) = 34.2 uC

Voltage on C1 (right) = Q1/C1 = 34.2/11.40 = 3.00 V

Voltage on C1 (left) = Q1/C1 = 34.2/11.40 = 3.00 V

Voltage on Cp = Qp/Cp = 34.2/11.40 = 3.00 V

Since C2 and 6.00 uF are in parallel, So

Voltage on C2 = Voltage on 6.00 uF = Vp

Voltage on C2 = 3.00 V

Charge on C2 = C2*V2 = 5.40*3 = 16.2 uC

Voltage on 6.00 uF = 3.00 V

Charge on 6.00 uF = C*V = 6*3 = 18.0 uC

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