Question
A current of $$3\,A$$ flows through the $$2\,\Omega $$ resistor shown in the circuit. The power dissipated in the $$5\,\Omega $$ resistor is
A current of $$3\,A$$ flows through the $$2\,\Omega $$ resistor shown in the circuit. The power dissipated in the $$5\,\Omega $$ resistor is
A.
$$4\,W$$
B.
$$2\,W$$
C.
$$1\,W$$
D.
$$5\,W$$
Answer :
$$5\,W$$
Solution :
Voltage across $$2\,\Omega ,$$
$$ = 3 \times 2 = 6\,V$$
Voltage across $$4\,\Omega $$ and $$\left( {5\,\Omega + 1\,\Omega } \right)$$ resistor is same.
So, current across $$5\,\Omega = \frac{6}{{1 + 5}} = 1\,A\,\,\left[ {i = \frac{V}{R}} \right]$$
Power across $$5\,\Omega = P = {i^2}R$$
$$\eqalign{ & = {\left( 1 \right)^2} \times 5 \cr & = 5\,W \cr} $$
Voltage across $$2\,\Omega ,$$
$$ = 3 \times 2 = 6\,V$$
Voltage across $$4\,\Omega $$ and $$\left( {5\,\Omega + 1\,\Omega } \right)$$ resistor is same.
So, current across $$5\,\Omega = \frac{6}{{1 + 5}} = 1\,A\,\,\left[ {i = \frac{V}{R}} \right]$$
Power across $$5\,\Omega = P = {i^2}R$$
$$\eqalign{ & = {\left( 1 \right)^2} \times 5 \cr & = 5\,W \cr} $$
