In the figure R1 840 Ohm R2 R3 510 Ohm R4 898 Ohm and th

In the figure R_1 = 84.0 Ohm, R_2 = R_3 = 51.0 Ohm, R_4 = 89.8 Ohm, and the ideal battery has emf epsilon = 6.00 V. What is the equivalent resistance? What is i in resistance 1, resistance 2, resistance 3, and resistance 4?

Solution

a)

R2 ,R3 and R4 are in parallel ,so

1/R₂₃₄ = 1/51 + 1/51 + 1/89.8

R₂₃₄=19.86 ohms

R1 and R234 are in series ,so equivalent resistance is

R_eq=84+19.86 =103.86 ohms

b)

Total Current flowing in the circuit is

I=V/R_eq=6/103.86=0.05777A

So Current through Resistor 1 is

I₁=I=0.05777A

c)

Voltage acorss R₂₃₄ is

V₂₃₄=V*(R₂₃₄/R₁+R₂₃₄)=6*(19.86/19.86+84) =1.147 Volts

Current across R2 is

I₂=V₂/R₂ =1.147/51 =0.0225 A

d)

Current across R3 is

I₃=V₃/R₃ =1.147/51 =0.0225 A

e)

Current across R4 is

I₄=V₄/R₄ =1.147/89.8 =0.01277 A