Question

A cell can supply currents of \(1\ \text{A}\) and \(0.5\ \text{A}\) via resistances of \(2.5\ \Omega\) and \(10\ \Omega\) respectively. The internal resistance of the cell is

• A. \(2\ \Omega\)
• B. \(3\ \Omega\)
• C. \(4\ \Omega\)
• D. \(5\ \Omega\)

Hint:

For a cell supplying current through resistance \(R\), always use \(I=\dfrac{E}{R+r}\), where \(r\) is the internal resistance of the cell.

Answer 1

The Correct Option is D

Step 1: Use the relation for current supplied by a cell.
If \(E\) is the emf and \(r\) is the internal resistance, then
\[ I=\frac{E}{R+r} \]
where \(R\) is the external resistance.

Step 2: Form equations using the given data.
When external resistance is
\[ R_1=2.5\ \Omega \] the current is
\[ I_1=1\ \text{A} \]
So,
\[ 1=\frac{E}{2.5+r} \]
\[ E=2.5+r \]
Now, when external resistance is
\[ R_2=10\ \Omega \] the current is
\[ I_2=0.5\ \text{A} \]
Thus,
\[ 0.5=\frac{E}{10+r} \]
\[ E=\frac{10+r}{2} \]

Step 3: Equate the two expressions for emf.
\[ 2.5+r=\frac{10+r}{2} \]
Multiply by \(2\):
\[ 5+2r=10+r \]
\[ r=5\ \Omega \]

Step 4: Final conclusion.
Hence, the internal resistance of the cell is
\[ \boxed{5\ \Omega} \]