Question

Construct the following circuit in the simulation by adding a third resistor in parallel with resistor 2. R3 w w R R2 Choose the following:R, 259, R2 = 509, R3 = 1000, V = 5V 17. What is the voltage measured across R,? (V) 18. What is the voltage measured across R2? (V) 19. What is the voltage measured across R3? (V) 20. Using Ohm's law what is the current flowing through R,? (A) 21. What is the current through R2? (A) 22. What is the current through R? (A) 23. What is the equivalent resistance R23 of the parallel combination of R2 and R3? (2) 24. What is the equivalent resistance Req of the whole system? i.e. the series combination of Ry with R23? (2) 25. What is the current flowing through the equivalent resistor R23? (A) 26. What is the current flowing through the equivalent resistance Req in this case? i.e. the total current? (A)

Answer 1

23.

Equivalent resistor of 2 and 3 R₂₃ = R₂R₃ ÷ (R₂+R₃) = (50×100) ÷ (50+100) = 5000 ÷ 150 = 33.33 ohm

24. equivalent resistance of corcuit R_eq = R₁ + R₂₃ =

25 + 33.33 = 58.33 ohm

25. And 26.    R₂₃ and R₁ connect in series so equal current flow through both , and it is equal to total current.

I_eq = V_eq ÷R_eq 5 ÷ 58.33 = 0.0857 amp.

21 & 22 .

and R are in Parallel R2 In $0 Iz x0.0857 -0.0286 Amp. Retra 150 X2 0-089A R2 0:0857 - 0.057 1 Amp 132 R3 R₂ R3 100 150 Wh

17 ,18 &19.

Potential across resistorR1

V₁= I_eq R₁ = 0.0857×25 = 2.1425Volt

Potential across resistor R2

V₂= I₂R₂ = 0.0286× 50 = 1.43 volt

Potential across resistor R3

V3 = I3R3 = 0.0571× 100 = 5.71 volt

Question 20.

Current through R₁ is equal to equivalent current

0.0857 amp