Question

The figure shows a circuit diagram with a 24 V battery and negligible internal resistance. Calculate : (a) the reading of the ammeter. (b) reading of V₁, V₂ and V₃.

Answer 1

\[\text{Resistance of }6\Omega\text{ and }3\Omega\text{ in parallel}\]

\[\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3}   \]

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

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

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

\[\therefore R_p = 2\Omega\]

\[\text{Resistance of }1.5\Omega\text{, }R_p\text{ and }8.5\Omega\text{ in series}\]

\[R = 1.5 + 2 + 8.5 = 12\Omega\]

\[\text{(a) }\therefore \text{Reading of the ammeter (I)} = \frac{V}{R} = \frac{24}{12} = 2\text{A}\]

\[\text{(b) Reading of }V_1 = I.R = 2 \times 1.5 = 3\text{V}\]

\[\text{Reading of }V_2 = I.R_p = 2 \times 2 = 4\text{V}\]

\[\text{Reading of }V_3 = I.R = 2 \times 8.5 = 17\text{V}\]