Question

In the figure ε₁ = 5.14 V, ε₂ = 12.7 V, R₁ = 94.2 Ω, R₂ = 185 Ω, and R₃ = 288 Ω. One point of the circuit is grounded (V = 0). What are the (a) size and (b) direction (up or down) of the current through resistance 1, the (c) size and (d) direction (left or right) of the current through resistance 2, and the (e) size and (f) direction of the current through resistance 3? (g) What is the electric potential at point A?

Answer 1

Use Kirchhoff’s loop rule and junction rule to form three equations and solve them to find the currents and their directions as shown below

a) Using loop rule in left loop, Isat en II RS 112 -5.14U t Iz Rzt In Ri=0 5.14V – ERI + 12:70 - - - - =) -5.140 + Iz(1852) + 11 (94.22) = 0 ---O Using loop rule in night loop, -12.7V + Ig Rz + IL RE = 0 = -12.7V + 13 (2883) + I, (94-22) = 0 ---- © Using KCL, I = I2+I3 ---- 0 from @q-0.0 and I Ij = 0.0391] A b) down c) Iz = 10.00785 A d) sight e) Iz = 0.0313A fţ left 3) VA -O - IL RI => VA = (0.0391 A)(94.23) = 13.68] v