Question
The current $$\left( i \right)$$ in the given circuit is
The current $$\left( i \right)$$ in the given circuit is
A.
$$1.6\,A$$
B.
$$2\,A$$
C.
$$0.32\,A$$
D.
$$3.2\,A$$
Answer :
$$2\,A$$
Solution :
In the given circuit, resistances $${R_B}$$ and $${R_C}$$ are in series order, so their effective resistance,
$$\eqalign{ & R' = {R_B} + {R_C} \cr & = 6 + 6 = 12\,\Omega \cr} $$
Now, $${R_A}$$ and $${R'}$$ are in parallel order, hence net resistance of the circuit
$$R = \frac{{R' \times {R_A}}}{{R' + {R_A}}} = \frac{{12 \times 3}}{{12 + 3}} = \frac{{36}}{{15}}\Omega $$
The current flowing in the circuit,
$$i = \frac{V}{R} = 4.8 \times \frac{{15}}{{36}} = 2\,A$$
In the given circuit, resistances $${R_B}$$ and $${R_C}$$ are in series order, so their effective resistance,
$$\eqalign{ & R' = {R_B} + {R_C} \cr & = 6 + 6 = 12\,\Omega \cr} $$
Now, $${R_A}$$ and $${R'}$$ are in parallel order, hence net resistance of the circuit
$$R = \frac{{R' \times {R_A}}}{{R' + {R_A}}} = \frac{{12 \times 3}}{{12 + 3}} = \frac{{36}}{{15}}\Omega $$
The current flowing in the circuit,
$$i = \frac{V}{R} = 4.8 \times \frac{{15}}{{36}} = 2\,A$$
