Question

Consider the circuit shown in the figure

• A. the current in the $5\Omega$ resistor is $2A$
• B. the current in the $5\Omega$ resistor is $1A$
• C. the potential difference $V_A-V_B$ is $10V$
• D. the potential difference $V_A-V_B$ is $5V$

Answer 1

Answer: (A)

the current in the $5\Omega$ resistor is $2A$

The current through the voltage source can be found by evaluating the equivalent resistance and applying ohm's law.

Equivalent resistance:-

solving from the right most resistor:-

equivalent of $3\Omega ,4\Omega ,3\Omega \left(e_1\right)$ $=3+4+3=10\Omega$

equivalent of $e_1,10\Omega \left(e_2\right)$ $=\frac{10}{2}=5\Omega$

equivalent of $e_2,3\Omega ,2\Omega \left(e_5\right)$

$=5+2+3=10\Omega$

equivalent of $e_3,10$ $e_4=\frac{10}{2}=5\Omega$

now the net equivalent resistance(R) $=5+5+4=14\Omega$

current through the source $=\frac{28}{14}A=2A$

So the current through 5$\Omega$ resistance $=2A$

$\Rightarrow$ current through next 3$\Omega$ $=1A$

current through next adjacent 3$\Omega$ $=0.5A$

On applying 'Kirchhoff's voltage law' from A to B:-

$V_A-(3+4+3)0.5-2\left(1\right)-V_B=0$

$V_A-V_B=7V$