Question

The figure shows a circuit diagram containing 12 cells, each of e.m.f. 1.5 V and internal resistance 0.25 Ω. Calculate :

1. Total internal resistance
2. Total e.m.f.
3. Total external resistance
4. Reading shown by the ammeter
5. Current in 12 Ω and 8 Ω resistors
6. p.d. across 2.2 Ω resistor
7. Drop in potential across the terminals of the cell.

Answer 1

(a) Internal resistance of 12 cells in series \[=12 \times 0.25\] \[=3\text{ Ω}\]

(b) e.m.f. of 12 cells in series \[=12 \times 1.5\] \[=18\text{ V}\]

(c) Resistance of the upper parallel branch = \(4\ \Omega + 8\ \Omega = 12\ \Omega\)

Resistance of the parallel combination (\(R_{p}\)):

\[ \frac{1}{\mathrm{R}_{p}} = \frac{1}{12} + \frac{1}{12}\]

\[ \frac{1}{\mathrm{R}_{p}} = \frac{2}{12}\]

\[ \frac{1}{\mathrm{R}_{p}} = \frac{1}{6} \]

\[ \therefore \mathrm{R}_{p} = 6 \ \Omega\]

Total external resistance (\(R\)) = Ammeter resistance + Parallel combination resistance + Remaining series resistor

\(R=0.8\ \Omega +6\ \Omega +2.2\ \Omega\)

\(=0.8+8.2\)

\(={9\ \Omega }\)

(d) Current recorded by ammeter \[I=\frac{E}{R+r}\] \[=\frac{18}{9+3}\] \[=1.5\text{ A}\]

(e) Current in 12 Ω and 8 Ω resistors

Potential difference across the parallel circuit (\(V_{p}\)) = \(\text{Total Current} \times R_p\)

\(V_{p}=1.5\text{\ A}\times 6\ \Omega =9\text{\ V}\)

Current flowing through the lower \(12\ \Omega\) resistor:

\[I_{\text{lower}} = \frac{V_{p}}{\text{Resistance}} = \frac{9\text{ V}}{12\ \Omega} = {0.75\text{ A}}\]

Current flowing through the upper branch containing the \(8\ \Omega\) resistor (Total branch resistance is \(4 + 8 = 12\ \Omega\)):

\[I_{\text{upper}} = \frac{V_{p}}{\text{Total Branch Resistance}} = \frac{9\text{ V}}{12\ \Omega} = {0.75\text{ A}}\]

(f) p.d. across \(2.2\ \Omega\) resistor (V) 

\(V=I\times R=1.5\text{\ A}\times 2.2\ \Omega ={3.30}\text{\ V}\)

(g) Drop in potential across the terminals of the cell \[E-V=Ir=1.5 \times 3\] \[=4.5\text{ V}\]