Question

A battery of emf $10volt$ and internal resistance $2\Omega$ is connected to an external resistance $8\Omega$ as shown in the figure:-

• A. Work done due to conservative electric field while a unit positive charge passes through battery from $Q$ to $P$ (along the arrow) is $8Joule$
• B. Work done due to conservative electric field while a unit positive charge passes through battery from $Q$ to $P$ (along the arrow) is $-8Joule$
• C. Work done due to conservative electric field while a unit positive charge passes through $8\Omega$ along the arrow is $-8Joule$
• D. Work done due to conservative electric field while a unit positive charge move from $Q$ to $P$ (along the arrow) is $10Joule$

Answer 1

The correct option is A Work done due to conservative electric field while a unit positive charge passes through battery from $Q$ to $P$ (along the arrow) is $8Joule$

The battery of $emf=10V$

resistance $=2\Omega$

external resistance $=8\Omega$

The direction of the current is $Q$ to $P$ hence the work done due to conservative electric field while a unit $(+Ve)$ charges passes through battery from $Q$ to $P$ a along the arrow

and now the work done $=VI$

$=\frac{10\times 210\times 10}{{10}^{-2}}=10-2=8$