Question

The emfs in the figure below are &1 = 7.00 V and E2 = 18.0 V. The resistances are R1 = 18.0 , R2 = 31.5 , R3 = 47.00, and R4 = 58.0 N. Find the magnitude of the current in each resistor when the switch is in the following states. w R$ 3 R w R3 (a) open 11 = A A 12 = 13 = A A (b) closed 11 А A 12 = 13 = 14 = A А

Answer 1

a)When the switch is open then no current will flow through the R4 resistor. And the other 3 resistors will be in series.

So, the equivalent resistance is= (18+31.5+47) = 96.5

The current through this three resistors will be = (18/96.5) = 0.1865 Amp

when switch is open:

I1	0.1865 Amp
I2	0.1865 Amp
I3	0.1865 Amp
I4	0 Amp

b) When the switch is closed there is two loop. The circuit will be like:

Applying Kirchoff's law in the loop bcdeb:

-(i1 ×R2)+e2 - (i1×R3)-(i2×R1) = 0

=> 31.5×i1 + 47×i1 +18×i2 = 18

=> 78.5×i1 + 18×i2 = 18..........i)

Applying KCL at the e point :

i1 = i2 + i3

=> i3 = i1 - i2

Applying Kirchoff's law in the loop abcdefa:

-(i1×R2) +e2 -(i1 ×R3) -e1 -(i3×R4) = 0

=> -(31.5×i1) +18 -(47×i1) -7 - (58×i3) = 0

=> 78.5×i1 + 58×i3 = 11

=> 78.5×i1 + 58(i1 - i2) = 11

=> 136.5×i1 - 58×i2 = 11..........ii)

By solving the equation i) and ii) we get:

i1 = 0.177 Amp

i2 = 0.227 Amp

i3 = -0.05 Amp

[The negative sign denotes that the current is in opposite of the supposed direction]

when swith is closed:

i1	0.227 Amp
i2	0.177 Amp
i3	0.177 Amp
i4	0.05 Amp

Please comment if you have any doubt and like if it helps.

Happy learning.