Capacitor — NEET UG practice

1. Recall the standard result for the energy lost when a charged capacitor is connected to an identical uncharged capacitor. When two capacitors $C_1$ and $C_2$ share charge, the energy lost is given by: $U_{loss}=\frac{1}{2}\cdot\frac{C_1C_2}{C_1+C_2}\cdot V^2$

2. Substitute the given values, $C_1=C_2=200\ pF=200\times10^{-12}F$ and $V=100\ V$. $U_{loss}=\frac{1}{2}\left(\frac{200\times200}{200+200}\right)\times10^{-12}\times(100)^2$

3. Simplify the coefficient inside the bracket first. $\frac{200\times200}{400}=100$

4. Substitute this back and multiply through to get the final energy loss. $U_{loss}=\frac{1}{2}\times100\times10^{-12}\times10^4=\frac{1}{2}\times10^{-6}=0.5\times10^{-6}\ J$

Hence, the electrostatic energy lost is $0.5\times10^{-6}\ J$, matching Option A.

1. From the figure, P and Q are connected in series with each other, and this series combination is connected in parallel with S across the battery of voltage $V$. Since P and Q are identical capacitors in series, the total voltage $V$ splits equally between them. $V_P=V_Q=\frac{V}{2}$ S, being directly across the battery, gets the full voltage. $V_S=V$

2. To find the energy in each capacitor, recall the formula for electrostatic energy stored in a capacitor. $U=\frac{1}{2}CV^2$

3. Applying this formula to P (which has voltage $V/2$ across it) gives $U_P$. $U_P=\frac{1}{2}C\left(\frac{V}{2}\right)^2=\frac{CV^2}{8}$

4. Similarly, find $U_Q$ and $U_S$ using the same formula with their respective voltages. $U_Q=\frac{1}{2}C\left(\frac{V}{2}\right)^2=\frac{CV^2}{8}$ $U_S=\frac{1}{2}CV^2$

5. Total energy stored in the system is the sum of the energies of all three capacitors. $U_T=U_P+U_Q+U_S=\frac{CV^2}{8}+\frac{CV^2}{8}+\frac{CV^2}{2}=\frac{3CV^2}{4}$

6. Finally, divide $U_P$ by $U_T$ to get the required ratio. $\frac{U_P}{U_T}=\frac{\frac{CV^2}{8}}{\frac{3CV^2}{4}}=\frac{1}{6}$

Hence, the answer is Option A, $\frac{1}{6}$.

1. Reduce the network by combining the capacitors according to the figure — the given branches formed by $C_1, C_2, C_3, C_4$ simplify into two identical parallel paths, each of equivalent value $2.5\ \mu F$. Since capacitors in parallel simply add up, combine these two branches. $C_{eq}=2.5\ \mu F+2.5\ \mu F=5\ \mu F$

2. Use the definition of capacitance, $Q=CV$, to find the total charge delivered by the battery. Given $C_{eq}=5\ \mu F$ and $V=50\ V$, $Q=C_{eq}V=5\times50=250\ \mu C$

3. Since the two branches are identical and connected in parallel, the total charge splits equally between them, and every capacitor lying along a series path within a branch carries the same charge. $q_1=q_2=q_3=q_4=\frac{250}{2}=125\ \mu C$

4. For $C_5$, apply $Q=CV$ directly using its own capacitance and the 50 V across it. $q_5=C_5V=2.5\times50=125\ \mu C$

Hence, the equivalent capacitance is $5\ \mu F$, and every capacitor carries a charge of 125 μC, matching Option B.