JEE MAIN - Physics (2003 - No. 12)
A $$3$$ volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The current $${\rm I}$$, in the circuit will be
$$1$$ $$A$$
$$1.5$$ $$A$$
$$2$$ $$A$$
$$1/3$$ $$A$$
Explanation
$${R_p} = {{3 \times 6} \over {3 + 6}} = {{18} \over 9} = 2\Omega $$
$$\therefore$$ $$V = IR$$
$$ \Rightarrow I = {V \over R} = {3 \over 2} = 1.5A$$
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