© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Determine the initial energy stored in the charged capacitor
A capacitor of capacitance 5\,\mu \text{F} is charged to 220\,\text{V} . The energy stored ( u_i ) in a capacitor is given by
u = \tfrac{1}{2}\,C\,V^2.
Thus the initial energy is:
u_i \;=\; \tfrac{1}{2} \times 5 \times 10^{-6}\,\text{F} \times (220\,\text{V})^2.
First compute C \times V^2 :
5 \times 10^{-6}\,\text{F} \times 220^2
\;=\; 5 \times 10^{-6} \times 48400
\;=\; 5 \times 48400 \times 10^{-6}
\;=\; 242000 \times 10^{-6}
\;=\; 0.242\,\text{J}.
Hence
u_i
= \tfrac{1}{2} \times 0.242
= 0.121\,\text{J}.
Step 2: Find the final common potential
After disconnecting from the supply, the 5\,\mu \text{F} capacitor is combined with an uncharged 2.5\,\mu \text{F} capacitor in such a way that they share charge (according to the given solution steps). The final common potential V_f (as used in the official solution) is obtained by charge conservation in a parallel-like combination:
V_f \;=\; \dfrac{C_1 V_1 + C_2 V_2}{C_1 + C_2}
\;=\; \dfrac{5 \times 220 + 2.5 \times 0}{5 + 2.5}
\;=\; 220 \times \dfrac{5}{7.5}
\;=\; 220 \times \dfrac{2}{3}
\;=\; 146.666...\,\text{V} \approx 146.67\,\text{V}.
Step 3: Calculate the final energy stored in the combined capacitors
Once the capacitors share charge, the total effective capacitance for this final arrangement (again following the official solution’s approach) is C_1 + C_2 = 5\,\mu\text{F} + 2.5\,\mu\text{F} = 7.5\,\mu\text{F} . The final energy u_f is thus
u_f \;=\; \tfrac{1}{2}\,(C_1 + C_2)\,V_f^2
\;=\; \tfrac{1}{2} \times 7.5 \times 10^{-6}\,\text{F} \times \bigl(146.67\,\text{V}\bigr)^2.
First compute (C_1 + C_2)\,V_f^2 :
7.5 \times 10^{-6}\,\text{F} \times (146.67)^2
\;\approx\; 7.5 \times 10^{-6} \times 21506.76
\;\approx\; 161300.7 \times 10^{-6}
\;=\; 0.1613\,\text{J}.
Hence
u_f \;=\; \tfrac{1}{2} \times 0.1613
= 0.08065\,\text{J} \approx 0.0807\,\text{J}.
Step 4: Compute the change in energy
The change in energy \Delta u during redistribution is
\Delta u \;=\; u_i \;-\; u_f
\;=\; 0.121\,\text{J} \;-\; 0.08065\,\text{J}
\;=\; 0.04035\,\text{J}.
This is approximately 4.035 \times 10^{-2}\,\text{J}.
Step 5: Relate the energy change to the given form and find X
According to the problem statement,
\Delta u = \dfrac{X}{100}\,\text{J}.
Since we have found
\Delta u \approx 0.04035\,\text{J},
we set
0.04035 = \dfrac{X}{100},
which gives
X = 0.04035 \times 100
= 4.035.
Rounding to the nearest integer,
X \approx 4.
Answer
The value of X to the nearest integer is
4.
