Question

a) Calculate the equivalent resistance of this circuit. (1) b) Calculate the current flowing through each resistor. (3) c) Calculate the power dissipated by resistor R3. (1) R2 R3 4012 2012 Ri 302 10 V

Answer 1

Givew Cruit R. (luor) R3 (2004) 3our lov 20 1900 R &R a since R and R are in series. & R' = R2 + R₂ = 410 +20=600 Now R and R, are in parrell i Rreet = R, XR1 - 60x30 60+30 Rreet = 200! (6) Now current in circut Rre 20. 10 = 0.5 amp. det i curant in R, and o n ini, will go in R' (R + R) tome, op and k' are in parrell you UR = VR - 4, XR, = (1 - 1) XR ! 1.

; x 30 = (0.5-1, ) x 60 2 - 1-21,-) 31 = 1 = 1/3 = 0.333amb and ca i-i, = 0.5-0.333 = 0.167 aup " R. and Rz are in series. Hencee (i-1) curent will flow in Both Resisten. = current in R, = 0.333 auß Li-i = currentin R2 and R₃ = 0.167 and - n dessor Power dissipited in R3 P = vi & P = Vzxl; and ... - 1 3 Rs | putting the value P = (0.167)² x 20 = 0.55778 Walt P₂ = 0.55778W
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