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# Two Capacitors of Unknown Capacitances C1 and C2 Are Connected First in Series and Then in Parallel Across a Battery of 100 V. - CBSE (Science) Class 12 - Physics

#### Question

Two capacitors of unknown capacitances C1 and C2 are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C1 and C2. Also calculate the charge on each capacitor in parallel combination.

#### Solution

When the capacitors are connected in parallel,

Equivalent capacitance, CP=C1+C2

The energy stored in the combination of the capacitorsE_P=1/2C_pV^2

=>E_P=1/2(C_1+C_2)(100^2)=0.25J

(C1+C2)=5×105      .....(i)

When the capacitors are connected in series,

Equivalent capacitanceC_S=(C_1C_2)/(C_1+C_2)

The energy stored in the combination of the capacitors

E_S=1/2C_SV^2

=>E_S=1/2(C_1C_2)/(C_1+C_2)(100)^2=0.045J

1/2(C_1C_2)/(5xx10^(-5))(100)^2=0.045J

C1C2=0.045×104×5×105×2=4.5×1010

(C1C2)2=(C1+C2)24C1C2

(C1C2)2=25×10104×4.5×1010=7×1010

=>(C_1-C_2)=sqrt(7xx10^(-10))=2.64xx10^(-5)

C1C2=2.64×105  .....(ii)

Solving (i) and (ii), we get

C1 = 35 μF and C2 = 15 μF

When the capacitors are connected in parallel, the charge on each of them can be obtained as follows:

Q1=C1V=35×106×100=35×104 C

Q2=C2V=15×106×100=15×104 C

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Solution Two Capacitors of Unknown Capacitances C1 and C2 Are Connected First in Series and Then in Parallel Across a Battery of 100 V. Concept: Combination of Capacitors.
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