Question on: WAEC Physics - 2006
a cell of e.m.f. 1.5V and internal resistance 1.0\(\Omega\) is connected to two resistor of resistance 2.0\(\Omega\) and 3.0\(\Omega\) in series. Calculate the current through the resistors
A
0.25A
B
0.30A
C
0.35A
D
0.50A
Ask EduPadi AI for a detailed answer
Correct Option: A
I = \(\frac{E}{R + r} = \frac{1.5}{5 + 1}\)
= \(\frac{1.5}{6} = 0.25A\)
= \(\frac{1.5}{6} = 0.25A\)
Add your answer
Please share this, thanks!
No responses