Question

In the circuit shown in the figure, VA = VB = 10.0 and R1 = R2 = R3 = R4 = Rs = 50.0. Current Is flows up through Rs, and I5 = 0.0500 A. Determine the current 14 through R4. Assume the direction of positive current flow is downward through R4. mm VA 14 = A Determine the voltage V2 across R2. mm V2 = V w Determine the power P3 dissipated by R3. R P3 = W

Answer 1

see the diagram :

. izn w R i, bo in OoOSA > Ry & rin مصممه e R, i

given :

I5 = 0.05 A

R1 = R2 =R3 =R4 = R5 = 50 ohm

VA = VB = 10 V

Let I4 be the current through R4,

I2 be the current through R2 ,

I3 be the curretn through R3,

I be the total current and

V2 be the voltage across R2.

a) Applying Loop Law on the loop abcda :

$-V_{B}+i_{5}R_{5}+i_{4}R_{4}=0$

$\Rightarrow -10+0.05*50+i_{4}*50=0$

$\Rightarrow i_{4}=\frac{10-0.05*50}{50} = 0.15A$ [answer]

b) Applying junction law at d :

i = i4 - i5 = 0.15 - 0.05 = 0.10 A

Applying Loop Law on the loop efgade :

$V_{A}-i_{2}R_{2}-i_{4}R_{4}-iR_{1}=0$

$\Rightarrow 10-i_{2}*50-0.15*50-0.1*50=0$

$\Rightarrow i_{2} = \frac{10-0.15*50-0.1*50}{50}=-0.05A$

-ve sign of i2 means the direction of current i2 is opposite to what I have assumed.

Voltage across R2 = i2*R2 = 0.05*50 = 2.5 V [answer]

c) Applying Loop Law on the loop ghbag :

$-i_{3}R_{3}+10+i_{2}R_{2}=0$

$\Rightarrow i_{3}=\frac{10+i_{2}R_{2}}{R_{3}} = \frac{10-0.05*50}{50}=0.15A$

Power dissipated by R3 = $i_{3}^{2}R_{3}=0.15^{2}*50=1.125W$ [answer]