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

View answer & explanation

Correct answer is A

I = \(\frac{E}{R + r} = \frac{1.5}{5 + 1}\)

= \(\frac{1.5}{6} = 0.25A\)

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