Question

Figure 8.14 shows a circuit diagram having a battery of 24 V and negligible internal resistance. Calculate the:

(a) reading of ammeter
(b) reading of $V_1,V_2$ and $V_3$.
[5 MARKS]

Answer 1

Each subpart: 2.5 Marks

Resistance of $6\Omega$ and $3\Omega$ are in parallel.
$\therefore$ Their equivalent resistance,
$\frac{1}{R_1}=\frac{1}{6}+\frac{1}{3}=\frac{1+2}{6}=\frac{1}{2}$
or $R_1=2\Omega$
Now, resistors of $1.5\Omega ,R_1$ and $8.5\Omega$ are in series.
$\therefore$ Total resistance = 1.5 + $R_1$ + 8.5
$=1.5+2+8.5=12\Omega$
(a) Current drawn from cell (ammeter reading), $I=\frac{V}{R}=\frac{24}{12}=2A$
(b) Potential difference across $1.5\Omega$ resistor, $V_1=IR=2\times 1.5=3V$
Potential difference across $R_1$ , $V_2=IR=2\times 2=4V$
Potential difference across $8.5\Omega$ resistor, $V_3=IR=2\times 8.5=17V$