JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 19)
An ideal cell of emf 10 V is connected in circuit
shown in figure. Each resistance is 2 $$\Omega $$. The
potential difference (in V) across the capacitor
when it is fully charged is ______.
_2nd_September_Evening_Slot_en_19_1.png)
Answer
8
Explanation
_2nd_September_Evening_Slot_en_19_2.png)
Capacitor is fully charged
So no current is there in branch ADB
$$ \therefore $$ Effective circuit of current flow :
_2nd_September_Evening_Slot_en_19_3.png)
Req = $$\left( {{{4 \times 2} \over {4 + 2}}} \right) + 2$$
Req = $${4 \over 3} + 2$$ = $${{10} \over 3}\Omega $$
i = $${{10} \over {{{10} \over 3}}}$$ = 3A
So potential different across AEB
= 2 × 1 + 2 × 3 = 8V
Hence potential difference across Capacitor = 8V
Comments (0)
