Question

Consider the circuit shown in the figure

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

Answer 1

The correct option is A The current in the $5\Omega$ resistor is $2\text{A}$

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 $3\Omega ,4\Omega ,3\Omega \text{in series is}\left(R_1\right)=3+4+3=10\Omega$
Equivalent of $R_1,10\Omega \text{in parallel is}\left(R_2\right)=\frac{10}{2}=5\Omega$
Equivalent of $R_2,3\Omega ,2\Omega \text{is}\left(R_3\right)=5+2+3=10\Omega$
Equivalent of $R_3,10\Omega \text{in parallel is}\left(R_4\right)=\frac{10}{2}=5\Omega$
Equivalent of $R_4,5\Omega ,4\Omega \text{is}\left(R_5\right)=5+5+4=14\Omega$

Current through the source $I=\frac{V}{R}$
$I=\frac{28}{14}=2A$

Current through $5\Omega$ is $2A$

Current through $3\Omega$ is $1A$
Current through adjacent $3\Omega$ is $0.5A$

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

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

$V_A-V_B=7V$