Question

A new flashlight cell of e.m.f. 1.5 volts gives a current of 15 amps, when connected directly to an ammeter of resistance$0.04 \Omega$. The internal resistance of cell is

• A. $0.04 \Omega$
• B. $0.06 \Omega$
• C. $0.10 \Omega$
• D. $10 \Omega$

Answer 1

Answer: (B)

$0.06 \Omega$

Answer 2

Let the internal resistance of cell be r, then

$15 = \frac { 1.5 } { 0.04 + r }$ ⇒ $r = 0.06 \Omega$