JAMB - Physics (1981 - No. 14)
A cell of e.m.f 2V and internal resistance 1\(\Omega\) supplies a current of 0.5 amps to a resistance whose value is
0.5\(\Omega\)
1\(\Omega\)
2\(\Omega\)
2.5\(\Omega\)
3\(\Omega\)
Giải thích
1 = \(\frac{E}{R + r}\)
\(\frac{2}{R + 1}\) = 0.5
0.5R = 2 - 0.5 = 1.5
R = \(\frac{1.5}{0.5}\)
= 3\(\Omega\)
\(\frac{2}{R + 1}\) = 0.5
0.5R = 2 - 0.5 = 1.5
R = \(\frac{1.5}{0.5}\)
= 3\(\Omega\)