Question

In the circuit given below, five resistances of \(10 \Omega\), \(40 \Omega\), \(20 \Omega\), \(30 \Omega\), and \(60 \Omega\) are connected as shown to a battery of 18 volts.
Total resistance of the circuit

Hint:

For resistors in parallel, use \( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \), and for resistors in series, add them directly.

Answer 1

- The \(10 \Omega\) and \(40 \Omega\) resistors are in parallel: \[ \frac{1}{R_1} = \frac{1}{10} + \frac{1}{40} = \frac{4+1}{40} = \frac{5}{40} \] \[ R_1 = \frac{40}{5} = 8 \Omega \] - The \(20 \Omega\) and \(30 \Omega\) resistors are in parallel: \[ \frac{1}{R_2} = \frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} \] \[ R_2 = \frac{60}{5} = 12 \Omega \] - The total resistance in parallel with the \(60 \Omega\) resistor: \[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{8} + \frac{1}{12} = \frac{3+2}{24} = \frac{5}{24} \] \[ R_{\text{total}} = \frac{24}{5} = 18 \Omega \]