An ideal cell of emf 10, V is connected in circuit shown in figure. Each resistance is 2, Omega . The potential difference (in V)

Q: 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

a $R _{1}$ to $R _{5} \rightarrow$ each $2 \Omega$ Cap. is fully charged So no current is there in branch $ADB$ Effective circuit of current flow $R_{e q}=\left(\frac{4 \times 2}{4+2}\right)+2$ $R _{ eq }=\frac{4}{3}+2=\frac{10}{3} \Omega$ $i=\frac{10}{10 / 3}=3 A$ So potential different across $AEB$ $\Rightarrow 2 \times 1+2 \times 3=8 V$ Hence potential difference across Capacitor $=\Delta V = V _{ AEB }=8 V$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

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

A$8$
B$10$
C$15$
D$25$

JEE MAIN 2020, Diffcult

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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