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.
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