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Question: A capacitor is charged by using battery in four steps. 1st it is charged to voltage $\frac{V_0}{4}$ ...

A capacitor is charged by using battery in four steps. 1st it is charged to voltage V04\frac{V_0}{4} and maintained for a charging time T >> RC. Then voltage is raised to V02\frac{V_0}{2} without discharging the capacitor and again maintained for time T >> RC. The process is repeated two more time to reach voltage V0V_0 & the capacitor is charged to final voltage V0V_0. If in this process total energy dissipated across resistance is 1NCV02\frac{1}{N}CV_0^2 where C is capacitance, calculate N.

Answer

8

Explanation

Solution

When a capacitor is charged from an initial voltage ViV_i to a final voltage VfV_f by connecting it to a battery of voltage VfV_f, the charge flowing from the battery is ΔQ=C(Vf−Vi)\Delta Q = C(V_f - V_i). The work done by the battery is W=VfΔQ=VfC(Vf−Vi)W = V_f \Delta Q = V_f C(V_f - V_i). The change in energy stored in the capacitor is ΔU=12CVf2−12CVi2\Delta U = \frac{1}{2}CV_f^2 - \frac{1}{2}CV_i^2. The energy dissipated in the resistance is ED=W−ΔUE_D = W - \Delta U.

In this problem, the capacitor is charged in four steps:

Step 1: From Vi=0V_i = 0 to V1=V04V_1 = \frac{V_0}{4} using a battery of voltage V1V_1.

  • Work done by battery: W1=V1(CV1−C⋅0)=CV12=C(V04)2=116CV02W_1 = V_1 (C V_1 - C \cdot 0) = C V_1^2 = C (\frac{V_0}{4})^2 = \frac{1}{16} C V_0^2.
  • Change in stored energy: ΔU1=12CV12−12C(0)2=12C(V04)2=132CV02\Delta U_1 = \frac{1}{2} C V_1^2 - \frac{1}{2} C (0)^2 = \frac{1}{2} C (\frac{V_0}{4})^2 = \frac{1}{32} C V_0^2.
  • Energy dissipated: ED1=W1−ΔU1=116CV02−132CV02=132CV02E_{D1} = W_1 - \Delta U_1 = \frac{1}{16} C V_0^2 - \frac{1}{32} C V_0^2 = \frac{1}{32} C V_0^2.

Step 2: From Vi=V1=V04V_i = V_1 = \frac{V_0}{4} to V2=V02V_2 = \frac{V_0}{2} using a battery of voltage V2V_2.

  • Work done by battery: W2=V2(CV2−CV1)=V02(CV02−CV04)=V02(CV04)=18CV02W_2 = V_2 (C V_2 - C V_1) = \frac{V_0}{2} (C \frac{V_0}{2} - C \frac{V_0}{4}) = \frac{V_0}{2} (C \frac{V_0}{4}) = \frac{1}{8} C V_0^2.
  • Change in stored energy: ΔU2=12CV22−12CV12=12C(V02)2−12C(V04)2=12C(V024−V0216)=12C(3V0216)=332CV02\Delta U_2 = \frac{1}{2} C V_2^2 - \frac{1}{2} C V_1^2 = \frac{1}{2} C (\frac{V_0}{2})^2 - \frac{1}{2} C (\frac{V_0}{4})^2 = \frac{1}{2} C (\frac{V_0^2}{4} - \frac{V_0^2}{16}) = \frac{1}{2} C (\frac{3V_0^2}{16}) = \frac{3}{32} C V_0^2.
  • Energy dissipated: ED2=W2−ΔU2=18CV02−332CV02=432CV02−332CV02=132CV02E_{D2} = W_2 - \Delta U_2 = \frac{1}{8} C V_0^2 - \frac{3}{32} C V_0^2 = \frac{4}{32} C V_0^2 - \frac{3}{32} C V_0^2 = \frac{1}{32} C V_0^2.

Step 3: From Vi=V2=V02V_i = V_2 = \frac{V_0}{2} to V3=3V04V_3 = \frac{3V_0}{4} using a battery of voltage V3V_3.

  • Work done by battery: W3=V3(CV3−CV2)=3V04(C3V04−CV02)=3V04(CV04)=316CV02W_3 = V_3 (C V_3 - C V_2) = \frac{3V_0}{4} (C \frac{3V_0}{4} - C \frac{V_0}{2}) = \frac{3V_0}{4} (C \frac{V_0}{4}) = \frac{3}{16} C V_0^2.
  • Change in stored energy: ΔU3=12CV32−12CV22=12C(3V04)2−12C(V02)2=12C(9V0216−V024)=12C(5V0216)=532CV02\Delta U_3 = \frac{1}{2} C V_3^2 - \frac{1}{2} C V_2^2 = \frac{1}{2} C (\frac{3V_0}{4})^2 - \frac{1}{2} C (\frac{V_0}{2})^2 = \frac{1}{2} C (\frac{9V_0^2}{16} - \frac{V_0^2}{4}) = \frac{1}{2} C (\frac{5V_0^2}{16}) = \frac{5}{32} C V_0^2.
  • Energy dissipated: ED3=W3−ΔU3=316CV02−532CV02=632CV02−532CV02=132CV02E_{D3} = W_3 - \Delta U_3 = \frac{3}{16} C V_0^2 - \frac{5}{32} C V_0^2 = \frac{6}{32} C V_0^2 - \frac{5}{32} C V_0^2 = \frac{1}{32} C V_0^2.

Step 4: From Vi=V3=3V04V_i = V_3 = \frac{3V_0}{4} to V4=V0V_4 = V_0 using a battery of voltage V4V_4.

  • Work done by battery: W4=V4(CV4−CV3)=V0(CV0−C3V04)=V0(CV04)=14CV02W_4 = V_4 (C V_4 - C V_3) = V_0 (C V_0 - C \frac{3V_0}{4}) = V_0 (C \frac{V_0}{4}) = \frac{1}{4} C V_0^2.
  • Change in stored energy: ΔU4=12CV42−12CV32=12C(V0)2−12C(3V04)2=12C(V02−9V0216)=12C(7V0216)=732CV02\Delta U_4 = \frac{1}{2} C V_4^2 - \frac{1}{2} C V_3^2 = \frac{1}{2} C (V_0)^2 - \frac{1}{2} C (\frac{3V_0}{4})^2 = \frac{1}{2} C (V_0^2 - \frac{9V_0^2}{16}) = \frac{1}{2} C (\frac{7V_0^2}{16}) = \frac{7}{32} C V_0^2.
  • Energy dissipated: ED4=W4−ΔU4=14CV02−732CV02=832CV02−732CV02=132CV02E_{D4} = W_4 - \Delta U_4 = \frac{1}{4} C V_0^2 - \frac{7}{32} C V_0^2 = \frac{8}{32} C V_0^2 - \frac{7}{32} C V_0^2 = \frac{1}{32} C V_0^2.

The total energy dissipated across the resistance is the sum of the energy dissipated in each step: ED,total=ED1+ED2+ED3+ED4=132CV02+132CV02+132CV02+132CV02=4×132CV02=432CV02=18CV02E_{D,total} = E_{D1} + E_{D2} + E_{D3} + E_{D4} = \frac{1}{32} C V_0^2 + \frac{1}{32} C V_0^2 + \frac{1}{32} C V_0^2 + \frac{1}{32} C V_0^2 = 4 \times \frac{1}{32} C V_0^2 = \frac{4}{32} C V_0^2 = \frac{1}{8} C V_0^2.

The problem states that the total energy dissipated is 1NCV02\frac{1}{N}CV_0^2. Comparing this with the calculated total energy dissipated, we have: 1NCV02=18CV02\frac{1}{N}CV_0^2 = \frac{1}{8}CV_0^2 1N=18\frac{1}{N} = \frac{1}{8} N=8N = 8.