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2. Electric potential and capacitance — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics2. Electric potential and capacitance1 Mark
MCQ
In the figure below, what is the potential difference between the point $A$ and $B$ and between $B$ and $C$ respectively in steady state
  • A
    ${V_{AB}} = {V_{BC}} = 100\,V$
  • B
    ${V_{AB}} = 75\,V,\;{V_{BC}} = 25\;V$
  • ✓
    ${V_{AB}} = 25\,V,\;{V_{BC}} = 75\;V$
  • D
    ${V_{AB}} = {V_{BC}} = 50\;V$
✓

Answer

Correct option: C.
${V_{AB}} = 25\,V,\;{V_{BC}} = 75\;V$
c
(c) ${C_{eq}} = \frac{{(3 + 3) \times (1 + 1)}}{{(3 + 3) + (1 + 1)}} + 1 = \left( {\frac{{6 \times 2}}{{6 + 2}}} \right)\, + 1$$ = \frac{5}{2}\,\mu F$
 $Q = C \times V = \frac{5}{2} \times 100 = 250\,\mu C$
Charge in $6\,\mu F$ branch $ = VC = \left( {\frac{{6 \times 2}}{{6 + 2}}} \right)\,100$$ = 150\,\mu C$
${V_{AB}} = \frac{{150}}{6} = 25\,V$ and ${V_{BC}} = 100 - {V_{AB}} = 75\,V$

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