Answer
a) $I_{1}=I_{2}=0.67A$, $I_{3}=1A$, $I_{4}=I_{5}=0.4A$
b) $V_{1}=6.7V$
$V_{2}=3.35V$
$V_{3}=10V$
$V_{4}=2V$
$V_{5}=8V$
c) $P_{1}=4.4W$
$P_{2}=2.2W$
$P_{3}=10W$
$P_{4}=0.8W$
$P_{5}=3.2W$
d) $P_{total}=21W$
Work Step by Step
a) Current in upper arm = $i_{1}=\frac{10}{R_{1}+R_{2}}=10/15=0.67A$
$i_{2}=10/10=1A$
$i_{3}=10/25=0.4A$
Thus, $I_{1}=I_{2}=0.67A$, $I_{3}=1A$, $I_{4}=I_{5}=0.4A$
b) $V_{1}=0.67\times 10=6.7V$
$V_{2}=0.67\times 5=3.35V$
$V_{3}=1\times 10=10V$
$V_{4}=0.4\times 5=2V$
$V_{5}=0.4\times 20=8V$
c) As , $P=I^{2}R$
Putting the corresponding values of i and R,
$P_{1}=4.4W$
$P_{2}=2.2W$
$P_{3}=10W$
$P_{4}=0.8W$
$P_{5}=3.2W$
d) $P_{total}=4.4+2.2+10+0.8+3.2=21W$
