Question

P4.9 For the circuit shown in Figure 4.14, R,= 102, Rz=522, Rz=812, R = 52, R = 10 , and the voltage across the network is 24 V. With the switch open, determine the a) equivalent resistance of the network, b) direction and magnitude of the current everywhere through the circuit and c) voltage at point "a" with respect to point “b”. R R; R W 21 +1 RE RA R, ŽR WW b) RE Figure 4.14 Problem 4.9 RA Figure 4.13 Problem 4.8 P4.10 For the circuit shown in Figure 4.14, use the same values as in Problem 4.9. With the switch closed, what is the a) current through the switch and b) equivalent resistance of the network.

P4.10 For the circuit shown in Figure 4.14, use the same values as in Problem 4.9. With the switch closed, what is the a) current through the switch and b) equivalent resistance of the network.  

Answer 1

P4.10. When the switch is closed, the three resistors, namely, R3, series combination of R1 and R4, series combination of R2,R5 - all these three will be in parallel with the battery of 24 V and voltage across each of them is therefore, 24V.

(a) Current through switch = current through R3

= voltage/R3

= (24/8) A = 3 A

(b) Eqivalent of R1,R4 in series, R' = R1+R4

= (10+5) ohm

= 15 ohm

Equivalent of R2,R5 in series, R'' = R2+R5

= (5+10) ohm

= 15 ohm

Finally, R3, R' and R'' all three are in parallel, so,

1/Req = (1/8 + 1/15 + 1/15) = (15+8+8/120)

Thus, equivalent resistance, Req = (120/31) ohm

= 3.87 ohm