Question

In the circuit diagram given below, five resistances of 10 \(\Omega\), 40 \(\Omega\), 30 \(\Omega\), 20 \(\Omega\), and 60 \(\Omega\) are connected as shown to a 12 volt battery. Calculate:

[(a)] Total (equivalent) resistance in the circuit.
[(b)] Total current flowing in the circuit.

Hint:

For parallel resistors: \[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots \] Add series resistances directly: \[ R_{\text{total}} = R_1 + R_2 + \ldots \]

Answer 1

(a) Total Equivalent Resistance
Step 1: Calculate equivalent resistance of the left parallel branch: \[ \frac{1}{R_1} = \frac{1}{10} + \frac{1}{40} = \frac{4 + 1}{40} = \frac{5}{40} = \frac{1}{8} \Rightarrow R_1 = 8\,\Omega \]
Step 2: Calculate equivalent resistance of the right parallel branch: \[ \frac{1}{R_2} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{2 + 3 + 1}{60} = \frac{6}{60} = \frac{1}{10} \Rightarrow R_2 = 10\,\Omega \]
Step 3: Total equivalent resistance in series: \[ R_{\text{total}} = R_1 + R_2 = 8 + 10 = 18\,\Omega \]
(b) Total Current Using Ohm’s Law: \[ I = \frac{V}{R} = \frac{12\,\text{V}}{18\,\Omega} = \frac{2}{3}\,\text{A} \approx 0.67\,\text{A} \]