Search any question & find its solution
Question:
Answered & Verified by Expert
In steady state, a capacitor of capacitance $2 \mu \mathrm{F}$ is charged to $4 \mu \mathrm{C}$, as shown in figure. If the internal resistance of the cell is $0.5 \Omega$, then the emf of the cell is
Options:
Solution:
1581 Upvotes
Verified Answer
The correct answer is:
$2.5 \mathrm{~V}$
$$
\mathrm{V}_C=\frac{Q}{C}=\frac{4 \times 10^{-6}}{2 \times 10^{-6}}=2 \mathrm{v}
$$
Now, $\mathrm{V}_C=\mathrm{V}$ (across $2 \Omega$ resistor)
$$
\therefore \quad I=\frac{V}{R}=\frac{2}{2}=1 \mathrm{~A}
$$
Now, using the relation, $\mathrm{V}=E-I r$
So, $\mathrm{E}=V+I r=2+1 \times 0.5=2.5 \mathrm{~V}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.