Question

A parallel combination of a 1.47 μ F 1.47 μF capacitor and a 2.97 μ F 2.97 μF capacitor is connected in series to a 4.51 μ F 4.51 μF capacitor. This three‑capacitor combination is connected to a 15.5 V 15.5 V battery. Find the charge of each capacitor.a) Charge of 4.25-μF capacitor: ______________ C
b) Charge of 1.13-μF capacitor: ______________ C
c) Charge of 2.85-μF capacitor: ______________ C

Answer 1

Answer:

The question is incomplete

the complete question is giving below

"parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capacitor combination is connected to a 17.3-V battery. Find the charge of each capacitor."

a. 35.6μC

b. 0.1μC

c. 38.0μC

Explanation:

Recall that for capacitors

C₁ and C₂ in parallel, the equivalent capacitance is given as

Cₙ=C₁+C₂

while for capacitors in series the equivalent capacitance is

Cₙ=1/C₁ +1/C₂

Hence for the parallel combination of 1.13μf and 2.85μf the equivalent capacitance is

Cₙ=1.13μf+2.85μf=3.98μf

Hence the 3.98μf is in series with 4.25μf. The equivalent of the series combination is

Cₙ=(4.25*3.98)/(4.25+3.98)μf

Cₙ=(16.915)/(8.23)μf

Cₙ=2.06μf

Next we determine the charge in the 2.06μf equivalent capacitor using the relationship

Q=CV

where V=17.3v

Q=17.3*2.06*10⁻⁶C

Q=35.6μC

a. Since the 3.98μf and then 4.25μf are in series, the charge value in the 4.25μf is 35.6μC

b. to determine the charge value, we solve for the voltage in the parallel combination which is

V=Q/C

V=35.6μC/3.98μf

V=8.94v

since the two capacitors are arranged in parallel, same voltage will be across them

Hence the charge on the 1.13μf is

Q=CV=1.13μf*8.94v

Q=10.1μC

c. the charge in the 4.25μf is

Q=4.25μf*8.94v

Q=38.0μC