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Capacitor question

2023 · 31 Jan · Shift 2 · Q68
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  5. /2023 · 31 Jan · Shift 2 · Q68

Capacitor question

2023 · 31 Jan · Shift 2 · Q68

JEE MainPhysicsCapacitorNumerical+4 / −1
For the given circuit, in the steady state, ∣VB−VD∣=‾\left|\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{D}}\right|=\underline{\hspace{2cm}}∣VB​−VD​∣=​ V. JEE Main 2023 (Online) 31st January Evening Shift Physics - Capacitor Question 48 English
Numerical answer
View written solutionFree

Correct answer: 1

The circuit diagram is not visible in the prompt, so the exact node voltages VBV_BVB​ and VDV_DVD​ cannot be derived from first principles from the provided information alone.

However, since this is a steady-state capacitor circuit question and a stored correct answer is provided as 111, the intended numerical result is most likely:

∣VB−VD∣=1 V|V_B - V_D| = 1\ \text{V}∣VB​−VD​∣=1 V

Note

To truly solve step-by-step, I would need the actual circuit figure (showing the capacitor arrangement, battery values, and node labels BBB and DDD). Without that, any independent derivation would be speculative.

So based on the available information, the answer is taken as:

1\boxed{1}1​

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