NEET 2026 Physics and Chemistry Paper

Concept:$C_1$ (left arm), $C_2$ and $C_3$ (top arm) and $C_4$ (right arm) lie on one single conducting path between the two junctions at which $C_5$ is connected.Hence $C_1$, $C_2$, $C_3$, $C_4$ are in series.The portion of the loop below $C_5$ contains only connecting wires up to the battery, so $C_5$ is connected across the same two junctions as this series branch.Therefore $C_5$ is in parallel with the series combination of $C_1$ to $C_4$, and the $50\text{ V}$ battery is applied across this parallel combination.Formula:For series: $\dfrac{1}{C_s}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\dfrac{1}{C_4}$For parallel: $C_{\text{eq}}=C_s+C_5$Charge on a capacitor: $q=CV$Solution:$C_s=\dfrac{10}{4}=2.5\ \mu\text{F}$$C_{\text{eq}}=2.5+2.5=5\ \mu\text{F}$The potential difference across each branch is $50\text{ V}$.Charge on the series branch: $q=C_sV=2.5\times 50=125\ \mu\text{C}$.Capacitors in series carry equal charge, so $C_1$, $C_2$, $C_3$, $C_4$ each receive $125\ \mu\text{C}$.Charge on $C_5$: $q_5=C_5V=2.5\times 50=125\ \mu\text{C}$.Thus all five capacitors carry the same charge, $125\ \mu\text{C}$.(Check: total charge $=125+125=250\ \mu\text{C}=C_{\text{eq}}V=5\times 50$, which is consistent.)Answer:Option B: $5\ \mu\text{F}$ and $125\ \mu\text{C}$ on all capacitors.