Question

In the given circuit, find: 
(i) Total resistance 
(ii) Current through the cell, and 
(iii) Potential difference across 0.8 \(\Omega\) resistance. 

Hint:

For resistors in series, the total resistance is the sum of individual resistances, and for resistors in parallel, the total resistance is the reciprocal of the sum of reciprocals of individual resistances.

Answer 1

In this circuit, we have resistances of \(6 \, \Omega\), \(2 \, \Omega\), and \(0.8 \, \Omega\) connected. To solve, we will first find the total resistance using the formula for resistors in series and parallel. First, the resistances \(6 \, \Omega\) and \(2 \, \Omega\) are in series: \[ R_{\text{total1}} = 6 + 2 = 8 \, \Omega \] Now, this \(8 \, \Omega\) is in parallel with \(0.8 \, \Omega\), so the total resistance is: \[ \frac{1}{R_{\text{total}}} = \frac{1}{8} + \frac{1}{0.8} = 0.125 + 1.25 = 1.375 \] \[ R_{\text{total}} = \frac{1}{1.375} \approx 0.727 \, \Omega \] Next, using Ohm's Law (\(V = IR\)), where the voltage is \(12 \, \text{V}\), we can find the current through the cell: \[ I = \frac{V}{R_{\text{total}}} = \frac{12}{0.727} \approx 16.5 \, \text{A} \] Now, for the potential difference across the \(0.8 \, \Omega\) resistance, we use Ohm's Law again: \[ V_{0.8 \, \Omega} = I \times 0.8 = 16.5 \times 0.8 \approx 13.2 \, \text{V} \]