A cell of emf 2 V and internal resistance 0.1 Ω is connected across a 3.9 Ω resistor. The current in the circuit is:
A20 A
B2.0 A
C0.5 A
D0.05 A
Answer & Solution
Correct answer: C. 0.5 A
I = ε/(R + r) = 2/(3.9 + 0.1) = 2/4 = 0.5 A.
Related questions
For a battery of emf ε and internal resistance r driving current I through external resistA galvanometer of resistance G is converted into a voltmeter reading full-scale voltage V A galvanometer of resistance G shows full-scale deflection at current I_g. To convert it iThe relation between current density j, drift velocity v_d, number density n and charge e A wire has a resistance R. It is stretched uniformly so that its length becomes 2L. The neKirchhoff current law at a junction is a statement ofTwo cells of emf ε and internal resistance r each are connected in parallel. The equivalenA potentiometer of length L is used to compare emfs. If the balance lengths for cells of e