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 \[V=iR=1.2\times 4=4.8\,V\]You need to login to perform this action.
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