Question

When a resistor of 11$\Omega$ is connected in series with an electric cell, the current flowing in it is $0.5A$. Instead when a resistor of $5\Omega$ is connected to the same electric cell in series, the current increases by $0.4A$. The internal resistance of the cell is

• A. $1.5\Omega$
• B. $2\Omega$
• C. $2.5\Omega$
• D. $3.5\Omega$

Answer 1

The correct option is C $2.5\Omega$

Case 1 :-
$E=I(R+r)$
$E=0.5(11+r)$
Case 2:-
$E=I(R+r)$
$E=0.9(5+r)$
$\Rightarrow 5(11+r)=9(5+r)$
$55+5r=45+9r$
$4r=10$
$r=2.5\Omega$