Question 74: Current Electricity

Question

When an external resistance of $5 \Omega$ is connected across terminals of a cell, a current of 0.25 A flows through it. When the $5 \Omega$ resistor is replaced by a $2 \Omega$ resistor, a current of 0.5 A flows through it. The internal resistance of the cell is $\_\_\_\_$ $\Omega$.

Accepted Answer

Let the emf of the cell be $E$ and its internal resistance be $r$.
Using OhmÃ¢ÂÂs law for the complete circuit,
$$ E = I(R+r) $$
First case
External resistance $R = 5\Omega$, current $I = 0.25	ext{ A}$
So,
$$ E = 0.25(5+r) $$
Second case
External resistance $R = 2\Omega$, current $I = 0.5	ext{ A}$
So,
$$ E = 0.5(2+r) $$
Since $E$ is same in both cases,
$$ 0.25(5+r) = 0.5(2+r) $$
Multiply both sides by $4$:
$$ 5+r = 2(2+r) $$
$$ 5+r = 4+2r $$
$$ 5-4 = 2r-r $$
$$ r = 1\Omega $$
So, the internal resistance of the cell is
$$ \boxed{1\Omega} $$

Current Electricity

When an external resistance of 5 \Omega is connected across terminals of a cell, a current of 0.25 A flows through it. When the 5 \Omega resistor is replaced by a 2 \Omega resistor, a current of 0.5 A flows through it. The internal resistance of the cell is \_\_\_\_ \Omega.

Integer / Numeric Answer Type