A capacitor $C_1=6 \mu \mathrm{~F}$, initially charged with a cell of emf 5 V is disconnected and connected to another capacitor $C_2=12 \mu \mathrm{~F}$ which is initially neutral. The charges on $C_1$ and $C_2$ after connection are (Shift - II Memory Based)

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Correct answer: b) $10 \mu \mathrm{C}, 20 \mu \mathrm{C}$. Step 1: Initial Charge on ${C}_{1}$​ The charge on a capacitor is given by: $Q=CV$ Given: ${C}_{1}=6\mu F$ $V=5V$ ${Q}_{1}=(6\times 1{0}^{−6}F)\times (5V)=30\mu C$ Since ${C}_{2}$​ is initially uncharged , its initial charge is: ${Q}_{2}=0\mu C$ Step 2: Charge Conservation When the two capacitors are connected, charge redistributes while maintaining charge conservation: ${Q}_{\text{total}}={Q}_{1}+{Q}_{2}=30\mu C+0=30\mu C$ Step 3: Common Voltage after Connection Since the capacitors are connected in parallel , they share a common voltage ${V}_{f}$: ${V}_{f}=\frac{{Q}_{\text{total}}}{{C}_{1}+{C}_{2}}$ ${V}_{f}=\frac{30\mu C}{(6+12)\mu F}$V ​ ${V}_{f}=\frac{30}{18}V=\frac{5}{3}V$ Step 4: Finding Final Charges Using $Q=CV$ Final charge on ${C}_{1}$: ${Q}_{1}^{′}={C}_{1}{V}_{f}=(6\mu F)\times \frac{5}{3}V$ ${Q}_{1}^{′}=10\mu C$ Final charge on ${C}_{2}$: ${Q}_{2}^{′}={C}_{2}{V}_{f}=(12\mu F)\times \frac{5}{3}V$ ${Q}_{2}^{′}=20\mu C$ Step 5: Selecting the Correct Option The charges on \({C}

A capacitor \(C_1=6 \mu \mathrm{~F}\), initially charged with a cell of emf 5 V is disconnected and connected to another…

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