<div>QUESTION&nbsp;</div> <div><img src="https://images.topperlearning.com/topper/tinymce/imagemanager/files/b15a7c2d765c45aaa953f88e5a5afcff59d1c9a8b06621.26402603A.jpg" alt="" /></div>

Asked by gpnkumar0 | 2nd Oct, 2017, 10:37: AM

Expert Answer:

The resistors R2 and R5 are connected in series and the combination is connected in parallel with series combination of R3 and R4.   Same connection is made between R6, R9 and R7 and R8.   Now, series equivalent is R2 + R5 = R3 + R4 = 2 kΩ The parallel combination gives RP1 = 1 kΩ   Again R6 + R9 = R7 + R8 = 2 kΩ And parallel combination gives RP2 = 1 kΩ   Now, by analysing the circuit further R1, RP1 and RP2 are all in series. Hence, the total resistance of the circuit is   R = R1 + RP1 + RP2 = 1 + 1 + 1 = 3 kΩ   Therefore, the current through the circuit is   I = V/R = 6/3 (V/kΩ) = 2 mA

Aswered by Romal Bhansali | 25th Oct, 2017, 03:54: PM

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