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Saturday, 31 October

JEE Main 2020 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 2 September 2020, Shift 2, Physics Q24

Question 24 of 75 in this shift, Physics question 24 of 25.

An ideal cell of emf 10 V is connected in circuit shown in figure. Each resistance is 2 Ω\Omega. The potential difference (in V) across the capacitor when it is fully charged is __________.

The figure, in words

Circuit drawn as a rectangle. Top edge: resistor R1 at the left, then a junction, then a capacitor C, then to the right edge. Right edge: resistor R5 running down from the top-right corner to the node where R4 meets the right edge; the right edge then continues down to the bottom edge. Middle: resistor R2 drawn vertically from the junction after R1 down to a node on the horizontal middle line; the middle line runs from the left edge through resistor R3 (an arrow pointing right just before it) to that node, then through resistor R4 to the right edge below R5. Left edge has two upward arrows. Bottom edge has a cell labelled 10 V.

Official answer

8.00

NTA final key (Sep 2020).