Observe the circuit diagram given below.
Calculate the current:
(a) through 3Ω resistor.
(b) drawn from the cell when the 3Ω resistance is replaced by a connecting wire of negligible resistance.
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Given,
- E.m.f. of the cell (ε) = 6 V
- Internal resistance of the cell (r) = 0.6 Ω
- Resistances in parallel = 6 Ω and 4 Ω
- Resistance in series = 3 Ω
Let the resultant of the two parallel resistances be RP.
$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{6} + \frac{1}{4} \\[1em] \Rightarrow \frac{1}{\text R_\text P} = \frac{2 + 3}{12} = \frac{5}{12} \\[1em] \Rightarrow \text R_\text P = \frac{12}{5} = 2.4\ \text{Ω}\end{aligned}$$
(a) Now RP is in series with the 3 Ω resistor and the internal resistance of the cell.
Total resistance of the circuit (RS) = 2.4 + 3 + 0.6 = 6 Ω
Current drawn from the cell = $\frac{\text ε}{\text R_\text S} = \frac{6}{6}$ = 1 A
Since the 3 Ω resistor is in series with the cell, the whole of this current passes through it.
Hence, the current through the 3 Ω resistor is 1 A.
(b) When the 3 Ω resistance is replaced by a connecting wire of negligible resistance,
Total resistance of the circuit = 2.4 + 0.6 = 3 Ω
Current drawn from the cell = $\frac{\text ε}{\text R} = \frac{6}{3}$ = 2 A
Hence, the current drawn from the cell is 2 A.