Question

A cell supplies a current of $1A$ through a resistance of $5\Omega$ and a current of $0.6A$ through a resistance of $10\Omega$. What is the internal resistance of the cell?

• A. $1.0\Omega$
• B. $1.5\Omega$
• C. $2.0\Omega$
• D. $2.5\Omega$.

Answer 1

The correct option is D $2.5\Omega$.

Given:

$I=1A$ and $R=5\Omega$

So, $\epsilon =I\left(R+r\right)$ ....($1$)

also $I^{\prime }=0.6A$ and $R^{\prime }=10\Omega$

So, $\epsilon =I^{\prime }\left(R^{\prime }+r\right)$ ....($2$)

Both equation ($1$) and ($2$) are equal, therefore

$\begin{array}{c}I\left(R+r\right)=I^{\prime }\left(R^{\prime }+r\right)\\ IR+Ir=I^{\prime }{R}^{\prime }+I^{\prime }r\\ r=\frac{I^{\prime }{R}^{\prime }-IR}{I-I^{\prime }}\\ r=\frac{0.6\times 10-1\times 5}{1-0.6}=\frac{10}{4}\\ r=2.5\Omega \end{array}$