Q175 / 180 · COMEDK 2024 (Morning Shift) · Physics

A cell of emf E and internal resistance r is connected to two external resistances $\mathrm{R_1}$ and $\mathrm{R_2}$ and a perfect ammeter. The current in the circuit is measured in four different situations: (a) without any external resistance in the circuit. (b) with resistance $\mathrm{R_1}$ only (c) with $\mathrm{R_1}$ and $\mathrm{R_2}$ in series combination. (d) with $\mathrm{R_1}$ and $\mathrm{R_2}$ in parallel combination. The currents measured in the four cases in ascending order are

  • A
    a < d < b < c
  • B
    c < d < b < a
  • C
    a < b < d < c
  • Correct answer
    D
    c < b < d < a

Solution

The current $I$ in a circuit with a cell of emf $E$ and internal resistance $r$ connected to an external resistance $R_{ext}$ is given by $I = \dfrac{E}{r + R_{ext}}$.

The four cases for $R_{ext}$ are:
(a) $R_{ext} = 0$, so $I_a = \dfrac{E}{r}$.
(b) $R_{ext} = R_1$, so $I_b = \dfrac{E}{r + R_1}$.
(c) $R_{ext} = R_1 + R_2$, so $I_c = \dfrac{E}{r + R_1 + R_2}$.
(d) $R_{ext} = \dfrac{R_1 R_2}{R_1 + R_2}$, so $I_d = \dfrac{E}{r + \dfrac{R_1 R_2}{R_1 + R_2}}$.

Comparing the denominators:
Since $R_1 + R_2 > R_1 > \dfrac{R_1 R_2}{R_1 + R_2} > 0$, the denominators follow the order $r + R_1 + R_2 > r + R_1 > r + \dfrac{R_1 R_2}{R_1 + R_2} > r$.

Since the current is inversely proportional to the denominator, the currents follow the reverse order:
$\dfrac{E}{r + R_1 + R_2} < \dfrac{E}{r + R_1} < \dfrac{E}{r + \dfrac{R_1 R_2}{R_1 + R_2}} < \dfrac{E}{r}$.

Therefore, $I_c < I_b < I_d < I_a$.

Answer: c < b < d < a

Hence, the correct answer is option: 'D'.

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