356918
A car has a fresh storage battery of e.mf. 12 \(V\) and internal resistance \(2 \times {10^{ - 2}}\Omega \). If the starter motor draws a current of 80 \(A\). Then the terminal voltage when the starter is on is
1 \(10.4\,V\)
2 \(9.3\,V\)
3 \(12\,V\)
4 \(8.4\,V\)
Explanation:
Given, e.m.f \(e = 12V\), internal resistance \(r = 2 \times {10^{ - 2}}W,{\rm{current}}\,I = 80A\) The relation between the terminal voltage (\(V\)) and emf is \(V = e - Ir\) \(V = 12 - 80 \times 2 \times {10^{ - 2}}\) or \(V = 12 - 1.6 = 10.4V\)