Question

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Question (2a) For the network shown, determine the unknown quantities. Solution (2a): CLO4 (100%) [8 Marks)

Answer 1

Answer:- Note that, all resistors are in parallel, so all would have same voltage drop equal to the value of E.

From 5mA, 2mA goes through R3 and hence 3mA would be flowing through R2. Hence the voltage drop across R2 is-
= 0.003*4000 V
= 12 V.
Thus the value of E = 12 V.

The value of R3 is-

$R_{3}=\frac{12}{0.002}\ \Omega =6\ k\Omega$

The current through R1 is equal to (9 - 5) mA = 4 mA. Thus the value of R1 is-

$R_{1}=\frac{12}{0.004}\ \Omega =3\ k\Omega$

The total resistance is the parallel value of all the three resistance i.e.-

$\frac{1}{R_{T}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

$\Rightarrow \frac{1}{R_{T}}=\frac{1}{3000}+\frac{1}{4000}+\frac{1}{6000}$

$\boldsymbol{\Rightarrow R_{T}=1.33\ k\Omega }$