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
The reactance of a capacitor $C$ at frequency $f$ in an a.c. circuit is:A bird perched on a single high voltage wire is unharmed because:The two conservation laws embodied in Kirchhoff's rules are conservation of:The resistance offered by the cell itself is called:Which device detects the current in a Wheatstone bridge?In a Wheatstone bridge, the source is connected across the:The Wheatstone bridge is made up of how many resistors?At a junction, the current entering equals the current: