A) \[1.6\,\,V\]
B) \[2.4\,\,V\]
C) \[4.8\,\,V\]
D) \[9.6\,\,V\]
Correct Answer: C
Solution :
In such networks always begin with the resistor about which two information?s (out of\[R,\,\,V,\,\,I\]) are given, to calculate its third information. So, the voltage across the \[3\Omega \] resistor \[=3\times 0.8=2.4\]volts This is also the voltage across the \[6\Omega \] resistor connected in parallel. Now we know, two things about this resistor, so we will calculate its third variable. \[I=\frac{V}{R}=\frac{2.4}{6}=0.4\,\,Amp\] (In parallel branch of twice the resistance, the current can directly be said to be half). Now, the total current through the \[4\Omega \] resistor\[=0.8+0.4=1.2\,\,Amp\] Now, voltage across it,\[V=IR=4\times 1.2=4.8\]
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