question_answer

If the current through 3\[\Omega \] resistor is 0.8 A, then potential drop through 4 \[\Omega \] resistor is:

A) \[\text{1}\text{.2V}\]                                   

B)  \[\text{2}\text{.6V}\]

C)  \[\text{4}\text{.8V}\]                  

D)         \[\text{9}\text{.6V}\]

Correct Answer: C

Solution :

Key Idea: In parallel combination of two resistors potential drop across each is same. In the given circuit,\[3\,\Omega \] and \[6\,\Omega \] resistors are in parallel, so potential drops across them will be equal. Let \[{{i}_{1}}\] and \[{{i}_{2}}\]currents are flowing in \[3\,\Omega \]and \[6\,\Omega \] respectively.                                 \[3{{i}_{1}}=6{{i}_{2}}\]                 But \[{{i}_{1}}=0.8A\](given)                 \[\therefore \]  \[3\times 0.8=6{{i}_{2}}\]                 \[\Rightarrow \]               \[{{i}_{2}}=\frac{3\times 0.8}{6}=0.4A\] Net current through the circuit                                 \[i={{i}_{1}}+{{i}_{2}}\]                                 \[=0.8+0.4=1.2\,A\] This current will How through \[4\,\Omega \]resistor. So, potential drop across \[4\,\Omega \]resistor is             \[V=iR=1.2\times 4=4.8\,V\]