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Question

Two capacitors with capacitance values C1=2000±10pF and C2=3000±15pF are connected in series. The voltage applied across this combination is 𝑉 = 5.00 ± 0.02 V. The percentage error in the calculation of the energy stored in this combination of capacitors is _______.


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Solution

Step 1: Given data

It has been given that, the capacitance of capacitors C1and C2 are (2000±10)pF and (3000±15)pF connected in series. The voltage applied across this combination is V=(5±0.02)V.

We have to find the percentage error in the calculation of the energy stored in this combination of capacitors.

Step 2: Formula to be used

The energy stored in a capacitor or capacitors are,
E=12CV2
Here, E is the energy stored, C is the capacitance and V is the potential difference.

The charge is,

Q=CV

Where Q is the charge on capacitance.

For series combination, the resultant capacitance is,

1Cs=1C1+1C2

Here, Cs is the resultant capacitance.

Step 3: Find the percentage error in the calculation of the energy stored in this combination of capacitors

Let,

1Ceq=1C1+1C2 -------- 1

=2000×30005000

=1200pF

Step 4: Differentiate the equation

Now, differentiate equation 1. We get,

-dCeqCeq2=-dC1C12-dC2C22

dCeqCeq2=dC1C12+dC2C22

dCeq1200×10-122=10×10-122000×10-122+15×10-123000×10-122

dCeq=6pF

We know that,

U=12CeqV2

Here, U is energy stored and V is applied voltage.

So,

dUU×100=dCeqCeq+2dVV×100

=61200+2×0.025×100

=5×10-3+8×10-3×100

=0.013×100

=1.3%

Hence, two capacitors with capacitance values C1=2000±10pF and C2=3000±15pF are connected in series. The voltage applied across this combination is 𝑉 = 5.00 ± 0.02 V. The percentage error in the calculation of the energy stored in this combination of capacitors is 1.3%.


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