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STD 12 - 2. Electric potential and capacitance — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. Electric potential and capacitance4 Marks
MCQ
In the figure below, what is the potential difference between the points $A$ 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
Refer image $(1)$

capacitor $x$ and $y$ are in parallel.So thin capacitance will be

$C_{n e t}=3+3=6 u F$

Capacitors $m$ and $N$ are in parallel Son this net capacitance will be

$C_{n e t}=1+2 u F$

So, the above diagram counted into

(Refer image $2$)

at $t=0 i=0$

Now, For $V_{A B}$

(Refer image $3$)

$V_{A}+\frac{150 u c}{6 u F}=V_{B}$

$V_{A}-V_{b}=-25 v=\left|V_{A}-V_{B}\right|=25 v$

$V_{A B}=25 V$

For $V_{B C}$

$V_{B}+\frac{150 u c}{2 u F}=V_{c}$

$V_{B}-V_{c}=-75 V$

$\Rightarrow\left|V_{B}-V_{c}\right|=75 V \rightarrow V_{B C}=75$

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