Question

Study the diagram alongside carefully and answer the following:

1. Equivalent resistance between P and Q.
2. Reading shown by the ammeter.
3. Current flowing in the 3 Ω resistor.

Answer 1

Given Data:

Number of cells connected in series (n) = 2

e.m.f. of each cell (E) = 1.5 V

Internal resistance of each cell (r) = 0.5 Ω

Total e.m.f.

\[ \mathrm{E}_{total} = 2 \times 1.5 \]

\[ = 3 \ \text{V} \]

Total internal resistance

\[ r_{total} = 2 \times 0.5 \]

\[ = 1 \ \Omega \]

i. The 3 Ω and 6 Ω resistors are connected in parallel between points P and Q.

\[ \frac{1}{\mathrm{R}_{PQ}} = \frac{1}{3} + \frac{1}{6} \]

\[ = \frac{2 + 1}{6} \]

\[ = \frac{3}{6} \]

\[ = \frac{1}{2} \]

\[ \therefore \mathrm{R}_{PQ} = 2 \ \Omega \]

ii. The ammeter is in the main circuit, so it measures the total current.

\[ \mathrm{R}_{total} = \mathrm{R}_{PQ} + r_{total} \]

\[ = 2 + 1 \]

\[ = 3 \ \Omega \]

\[ \mathrm{I} = \frac{\mathrm{E}_{total}}{\mathrm{R}_{total}} \]

\[ = \frac{3}{3} \]

\[ = 1 \ \text{A} \]

iii. Potential difference across P and Q

\[ \mathrm{V}_{PQ} = \mathrm{I} \times \mathrm{R}_{PQ} \]

\[ = 1 \times 2 \]

\[ = 2 \ \text{V} \]

Current in the 3 Ω resistor

\[ \mathrm{I}_{3} = \frac{\mathrm{V}_{PQ}}{3} \]

\[ = \frac{2}{3} \]

\[ \approx 0.67 \ \text{A} \]