Question

Find the voltage \(V_{AB}\) and the current \(i_{BA}\) in the given circuit.

• A. \(24\,\text{V},\,2\,\text{A}\)
• B. \(24\,\text{V},\,1\,\text{A}\)
• C. \(18\,\text{V},\,2\,\text{A}\)
• D. \(18\,\text{V},\,1\,\text{A}\)

Answer 1

The Correct Option is B

Concept:
When several cells with internal resistances are connected in parallel, the equivalent emf and equivalent resistance are given by \[ E_{\text{eq}} = \frac{\sum \frac{E_i}{r_i}}{\sum \frac{1}{r_i}} \] \[ r_{\text{eq}} = \frac{1}{\sum \frac{1}{r_i}} \] This allows the complex network between \(A\) and \(B\) to be reduced to a single equivalent source. Step 1: Write the equivalent emf expression. \[ E_{\text{eq}} = \frac{\frac{27}{3}+\frac{27}{3}+\frac{27}{3}+\frac{0}{6}+\frac{27}{3}} {\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{3}} \] \[ E_{\text{eq}} = \frac{36}{9/6} \] \[ E_{\text{eq}} = 24\,\text{V} \] Thus, \[ V_{AB} = 24\,\text{V} \] Step 2: Find equivalent resistance. \[ r_{\text{eq}} = \frac{1} {\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{3}} \] \[ r_{\text{eq}} = \frac{2}{3}\,\Omega \] Step 3: Find current \(i_{BA\).} The external resistor between the equivalent source and point \(B\) is \(3\,\Omega\). Total resistance: \[ R_{\text{total}} = 3 + \frac{2}{3} \] Using Ohm's law: \[ i_{BA} = \frac{E_{\text{eq}}}{R_{\text{total}}} \] \[ i_{BA} = \frac{24}{3+\frac{2}{3}} \] \[ i_{BA} = \frac{24}{\frac{11}{3}} \] \[ i_{BA} \approx 1\,\text{A} \] Thus, \[ V_{AB} = 24\,\text{V}, \qquad i_{BA} = 1\,\text{A} \]