Question

What is the current through the resistor $6\Omega$ in the circuit shown in figure.

• A. 10/3 A
• B. 5/2 A
• C. 5/3 A
• D. 1/4 A

Answer 1

Answer: (C)

5/3 A

Let us first calculate the current flowing out of the voltage source.

The $4\Omega$ and $4\Omega$ resistors are in parallel, so their

equivalent resistance is $R_1=4\times 4/(4+4)=2\Omega$.

The $R_1=2\Omega$ and $4\Omega$ resistors are in series, so their

equivalent resistance is $R_2=2+4=6\Omega$.

The$2\Omega$ and $4\Omega$ in the rightmost branchare in series,

so their equivalent resistance is $R_3=4+2=6\Omega$

The $R_2=6\Omega$ and $R_3=6\Omega$are in parallel, so their

equivalent resistance is $R_4=6\times 6/(6+6)=3\Omega$

Now this $6\Omega$ and $3\Omega$, in upper-left part are in parallel,

so the equivalent resistance is $R_5=6\times 3/(6+3)=2\Omega$.

Finally, the $R_4=3\Omega$ and $R_5=2\Omega$ are in series so the

equivalent resistance is $R_6=3+2=5\Omega$

So the equivalent resistance of circuit is $5\Omega$.

And, current flowing out of voltage source $25V$ is $25/5=5A$

Now the current flowing through the $6\Omega$ resistor is

$5\times 3/(6+3)=5\times 3/9=5/3A$