Question

In the network shown, points $A,B$ and $C$ are at potentials of $70V$, zero and $10V$ respectively

• A. point $D$ is at a potential of $40V$
• B. the currents in the sections $AD,DB,DC$ are in the ratio $3:2:1$
• C. the currents in the sections $AD,DB,DC$ are in ration $1:2:3$
• D. the network draws a total power of $200W$

Answer 1

Answer: (A), (B), (D)

The correct options are
A point $D$ is at a potential of $40V$
B the currents in the sections $AD,DB,DC$ are in the ratio $3:2:1$
D the network draws a total power of $200W$

Redrawing and labeling given diagram as shown in figure.
Let V be the potential at D. Applying Kirchhoff's voltage law to branch AD, we get
$70-V=10i_1$
$-V=10i_1-70$......................(1)
Applying Kirchhoff's voltage law to branch BD, we get
$V-0=20i_2$
$V=20i_2$ or $i_2=\frac{V}{20}$ ......................(2)
Applying Kirchhoff's voltage law to branch CD, we get
$V-10=30(i_1-i_2)$.....................(3)
from (2) and (3)
$20i_2-10=30(i_1-i_2)$
$50i_2=30i_1+10$
$50\left(\frac{V}{20}\right)=30i_1+10$
$5V=60i_1+20$........................(4)
multiplying (1) by 5 and adding to (4), we get
$0=110i_1-330$
$i_1=\frac{330}{110}=3A$.................(5)
using (5) in (1)
$70-V=10\left(3\right)$
$V=40V$
Hence, point D is at a potential of 40 V.
from (2)
$40=20i_2$
$I_2=2A$
Current through branch DC is
$i_1-i_2=3-2=1$
Hence, the currents in the sections AD, DB, DC are in the ratio 3:2:1
Total power P, drawn by the network is
$P=(3{)}^{2}10+(2{)}^{2}20+(1{)}^{2}30$
$P=90+80+30$
$P=200W$