A cell of emf \( \varepsilon \) and internal resistance \( r \) is connected to an external resistance \( R \).
Given below are two statements:
Statement I: The maximum current that can be drawn from the cell is \( \varepsilon/r \), and it flows when \( R=0 \).
Statement II: If \( R \) is increased gradually from a very small value to a very large value, the power delivered to \( R \) increases continuously.
In the light of the above statements, choose the most appropriate answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Statement I is correct but Statement II is incorrect.
The emf of the cell drives the current through its internal resistance and the external resistance. These two are in series, so the current drawn from the cell is:
\( \displaystyle I=\frac{\varepsilon}{R+r} \)
Statement I The maximum current that can be drawn from the cell is \( \varepsilon/r \), and it flows when \( R=0 \). — True. The current is largest when the total resistance \( R+r \) is smallest. This happens when \( R=0 \):
\( \begin{aligned} I_{\max} &= \frac{\varepsilon}{0+r} \\ &= \frac{\varepsilon}{r} \end{aligned} \)
Statement II If \( R \) is increased gradually from a very small value to a very large value, the power delivered to \( R \) increases continuously. — False. The power delivered to the external resistance is:
\( \begin{aligned} P &= I^{2}R \\ &= \frac{\varepsilon^{2}R}{(R+r)^{2}} \end{aligned} \)
When \( R \) is very small, the current is close to \( \varepsilon/r \). But the factor \( R \) is nearly zero, so \( P \) is nearly zero.
When \( R \) is very large, the current \( \varepsilon/(R+r) \) becomes very small and \( P \) again falls towards zero.
So the power first rises and then falls as \( R \) is increased. It is largest for some in-between value of \( R \).
NCERT (2026-27):
Cells, Emf, Internal Resistance – “… we get”
\( IR=\varepsilon-Ir \)
\( \displaystyle \text{Or,}\ \ I=\frac{\varepsilon}{R+r} \qquad (3.40) \)
“The maximum current that can be drawn from a cell is for R = 0 and it is Imax = ε/r.”
Electrical Energy, Power – “… Using Ohm’s law V = IR, we get”
\( P=I^{2}R \)\( =V^{2}/R \qquad (3.33) \)
“as the power loss (‘ohmic loss’) in a conductor of resistance R carrying a current I.”