Question

A current of $2A$ flows through $2\Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5A$ when connected across a $9\Omega$ resistor. The internal resistance of the battery is

• A. $1/3\Omega$
• B. $1/4\Omega$
• C. $1\Omega$
• D. $0.5\Omega$

Answer 1

Answer: (A)

$1/3\Omega$

Given,

$I_1=2A,I_2=0.5A$

$R_1=2\Omega ,R_2=9\Omega$

Supplied voltage is same.

To find,

Internal Resistance of battery,$r=?$

$E=V+I\cdot r$

$E=I\cdot R+I\cdot r$

$E=I\cdot (R+r)$

$\therefore I=\frac{E}{(R+r)}$

$E_1=2\cdot (2+r)$ and $E_2=0.5\cdot (9+r)$

As the supplied voltage is same,

So, $E_1=E_2\Rightarrow 2\cdot (2+r)=0.5\cdot (9+r)$

$\Rightarrow \frac{2}{0.5}=\frac{9+r}{2+r}$

$\Rightarrow 4=\frac{9+r}{2+r}$

$\Rightarrow 4\cdot (2+r)=(9+r)$

$\Rightarrow 8+4r=9+r$

$\Rightarrow 4r-r=9-8$

$\Rightarrow 3r=1$

$\Rightarrow r=\frac{1}{3}\Omega$

Therefore, Internal Resistance of battery is $\frac{1}{3}\Omega$