## College Physics (7th Edition)

a). There would be no change in $I_{1}, I_{2}, I_{4}$. In the lower branch, the current will be halved since the new resistance is now doubled to 4Ohm. b). $I_{1} = I_{2}=3A$ and $I_{4}=6A$ Originally, $I_{3}=6A$ but now the new currents are $I'_{3}=I'_{5}=3A$
$R_{1}+R_{2}$, $R_{4}+R_{5}$, and $R_{3}$ are in parallel. Therefore $\frac{1}{Re}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}$ $Re=1 Ohm$ $i=12/1=12A$ Since potential across $R_{4}+R_{5}$ & $R_{3}$ are same, $i_{1}\times (R_{4}+R_{5})=(i-2i_{1})\times R_{3}$ $i_{1}=12/4=3A$ So, current in the upper arm = $3A$ and that in lower arm = $6A$ Originally, $I_{3}=6A$ but now the new currents are $I'_{3}=I'_{5}=3A$