Question
Current through $$3\,\Omega $$ resistor is $$0.8\,A,$$ then potential drop through $$4\,\Omega $$ resistor is
Current through $$3\,\Omega $$ resistor is $$0.8\,A,$$ then potential drop through $$4\,\Omega $$ resistor is
A.
$$9.6\,V$$
B.
$$2.6\,V$$
C.
$$4.8\,V$$
D.
$$1.2\,V$$
Answer :
$$4.8\,V$$
Solution :
Voltage across $$3\,\Omega $$ resistance is given by Ohm's law
i.e. $$V = IR = 3 \times {i_1}$$
$$ = 3 \times 0.8 = 2.4\,V\,\,\left( {{\text{as}}\,{i_1} = 0.8\,A} \right)$$
As $$3\,\Omega $$ and $$6\,\Omega $$ are in parallel, hence voltage across $$6\,\Omega $$ resistance will also be $$2.4\,V.$$
∴ Current through $$6\,\Omega $$ resistance
$${i_2} = \frac{V}{R} = \frac{{2.4}}{6} = 0.4\,A$$
∴ Total current in the circuit,
$$i = {i_1} + {i_2} = 0.8 + 0.4 = 1.2\,A$$
∴ Potential drop across $$4\,\Omega $$ resistance,
$$ = i \times 4 = 1.2 \times 4 = 4.8\,V$$
Voltage across $$3\,\Omega $$ resistance is given by Ohm's law
i.e. $$V = IR = 3 \times {i_1}$$
$$ = 3 \times 0.8 = 2.4\,V\,\,\left( {{\text{as}}\,{i_1} = 0.8\,A} \right)$$
As $$3\,\Omega $$ and $$6\,\Omega $$ are in parallel, hence voltage across $$6\,\Omega $$ resistance will also be $$2.4\,V.$$
∴ Current through $$6\,\Omega $$ resistance
$${i_2} = \frac{V}{R} = \frac{{2.4}}{6} = 0.4\,A$$
∴ Total current in the circuit,
$$i = {i_1} + {i_2} = 0.8 + 0.4 = 1.2\,A$$
∴ Potential drop across $$4\,\Omega $$ resistance,
$$ = i \times 4 = 1.2 \times 4 = 4.8\,V$$
