In the circuit diagram shown in the figure, the potential difference E is 20 V.
(A) The potential difference across points C and D is \( V_{CD} \)
(B) The potential difference across points B and C is \( V_{BC} \)
(C) The potential difference across points D and E is \( V_{DE} \)
(D) The potential difference across points A and B is \( V_{AB} \)
If the potentials \( V_{AB} \), \( V_{BC} \), \( V_{CD} \), \( V_{DE} \) are arranged in increasing order, then
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C), (A), (B), (D)
The four groups are in series with the cell, so the same current flows through all of them and the potential difference across each is \( V = I R \) with its own equivalent resistance. First find the resistance of each group:
\( R_{AB} = 4\,\Omega \)
\( R_{DE} = 1\,\Omega \)
\( \displaystyle R_{BC} = \frac{12 \times 4}{12 + 4} = 3\,\Omega \qquad R_{CD} = \frac{3 \times 6}{3 + 6} = 2\,\Omega \)
The total resistance and the current from the cell \( E = 20\,\mathrm{V} \) are
\( \displaystyle R_{\mathrm{total}} = 4 + 3 + 2 + 1 = 10\,\Omega \qquad I = \frac{20}{10} = 2\,\mathrm{A} \)
\( V_{AB} = 2 \times 4 = 8\,\mathrm{V} \)
\( V_{BC} = 2 \times 3 = 6\,\mathrm{V} \)
\( V_{CD} = 2 \times 2 = 4\,\mathrm{V} \)
\( V_{DE} = 2 \times 1 = 2\,\mathrm{V} \)
In increasing order, \( V_{DE} \lt V_{CD} \lt V_{BC} \lt V_{AB} \), that is (C), (A), (B), (D).
Shortcut: In a series circuit the potential difference across each group is proportional to its resistance (\( 1 : 2 : 3 : 4 \) for \( 1\,\Omega, 2\,\Omega, 3\,\Omega, 4\,\Omega \)), so the order of the potential differences is the order of the resistances.