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
