Question

a) For the Figures 1 and Figure 2 given below, find the current i (10 +10 Marks) 1) 0357 100T Figure 1 6 V Figure 2 b) In the network shown in Figure 3; (i) Let R = 802, find Reg (10 Marks) (ii) Find R if Req = 80 82; (10 Marks) 1012 40.2 w 0- M coppia - 31000 3300 3200

Answer 1

a) For fig 1 applying KVL in the loop

(voltage source for those current comig out of positive terminal is taken as positive and current going into positive terminal is taken as negative )

1-2 - 107 + 3.5 – 10i = 0

5 = 202

i = 0.125A

Similarly for fig 2 applying KVL in the loop

-10 - i+2-i-i-2 + 6- i=0

$\small -4i=4$

negative sign implies the direction of current is opposite to that of in the circuit

i=-1A

b) $\small (1)$ Given R=80 and we need to find equivalent resistance

Reg = ({{{40 + 20) ||30) +80) ||100)+10

Reg = (((60||30) +80) ||100)+10

Reg = (20 + 80) 100) + 10

Reg = (100 ||100)+10

Reg = 50+ 10

Req = 60

$\small (2)$ Given Equivalent resistance equal to 80 we need to find R

Reg = ({{{40 + 20) || 30) +R) || 100)+10

80 = (((60||30) + R) || 100)+10

80 = (20 + R) ||100) + 10

70 = (20 + R)||100)

70 = (20 + R) × 100/120 +R

70R + 8400 = 100R + 2000

30R = 6400

  

R=213.33

:) :) :)