Question

A $9V$ battery is connected to four resistor to form a simple circuit as shown in figure. How would the current through the $2\Omega$ resistor compare to the current through the $4\Omega$-ohm resistor?

• A. $One-fourth$
• B. $One-half$
• C. $Fourtimes$
• D. $Equal$

Answer 1

Answer: (D)

$R_{eq}=(7+2)//(4+5)=9//9=\frac{9\times 9}{9+9}=\frac{81}{18}=4.5\Omega$

$Current\left(I\right)=\frac{V}{R}=\frac{9}{4.5}=2A$

As both branch having equal resistance current will flow same in both branch

Therefore current flow in $7\Omega$ is $\frac{2}{2}=1A$

So equal current will follow in both resistance.