Question

Jn the 2-100f aircit It you er hnfed the JK-ohm resist t changes in olte drops and Current woul you expece7 What would harpen If You reversed one a the pawer suplies7 EXPlain R2=2kQ Ri 3kQ R3 1k A9-A V, 6V

Answer 1

Hi,

Hope you are doing well.

PART A:

Ri 3kQ R2 2k 12 2 R3 1k V, 6V A9-A

Applying Kirchhoff's junction rule,

131

Applying Kirchhoff's voltage rule in Loop 1 (left loop),

V1

R1(I112) R3= V1

300011I+ I2) X 1000 6 V

400011100012 6 (1)

Applying Kirchhoff's voltage rule in Loop 2 (right loop),

$I_{2}R_{2}+I_{3}R_{3}=V_{2}$

I2R2I1I) R3= Vz

I +I2) X 1000= 6 V 2000I2

1000113000I2 6 (2)

Solving equations (1) and (2), we get,

3 A 2750

and

$I_{2}=\frac{9}{5500}\;A$

3 A 13 I2- 100 I1I

Now voltage across 1k resistor is given by,

$V_{3}=I_{3}R_{3}=\frac{3}{1100}\;A\times1000\;\Omega$

$\mathbf{\therefore V_{3}=2.727\;V}$

PART B:

Lets reverse power supply on the right loop so that direction of I₂ will be reversed.

Applying Kirchhoff's junction rule,

3

Applying Kirchhoff's voltage rule in Loop 1 (left loop),

V1

$I_{1}R_{1}+\left (I_{1}-I_{2} \right )R_{3}=V_{1}$

$3000I_{1}+\left (I_{1}-I_{2} \right )\times1000=6\;V$

$4000I_{1}-1000I_{2} =6\rightarrow \left ( 1 \right )$

Applying Kirchhoff's voltage rule in Loop 2 (right loop),

$-I_{2}R_{2}+I_{3}R_{3}=V_{2}$

$-I_{2}R_{2}+\left (I_{1}-I_{2} \right )R_{3}=V_{2}$

$-2000I_{2}+\left (I_{1}-I_{2} \right )\times1000=6\;V$

$1000I_{1}-3000I_{2} =6\rightarrow \left ( 2 \right )$

Solving equations (1) and (2), we get,

3 A 2750

and

$I_{2}=\frac{-9}{5500}\;A$

$\therefore I_{3}=I_{1}+I_{2}=\frac{-3}{1100}\;A$

Now voltage across 1k resistor is given by,

V3I3R3 A x 1000 2 1100

$\mathbf{\therefore V_{3}=-2.727\;V}$

That is Magnitude of the current remains the same but, direction of I₂ and I₃ reverses.

Time elapsed: 50 min. 32 s.