Question

3. An electric circuit consists of a battery, three 2 12 resistors and a 10 A fuse. 12V 2Ω Calculate the total current out of the battery. If the current through the fuse exceeds 10 A, assume that it melts and opens the circuit. a) Determine the amount of charge that flows between the top two resistors at point B during a period of 5 seconds. b) c) Assume that the potential on the batteries positive terminal to be 12 V. Calculate the voltage at the following points: i) Point A ii) Point B ii) Point C d) Calculate the power dissipated through the single 2 Ω resistor in the lower branch.

Answer 1

In the above fig we have to determine R_eq 1st here two 2 ohm are in series and both are in parallel to 2 ohm so, 1/R_eq = 1/2 + 1/2 + 2 rightarrow_R_eq = 1.333 ohm So total current = epsilon/R_eq = 12 /1.33 = 9 pi [The fuse will not melt] Current flowing through ABC, I_B = I_roe times 2/2 + 2 + 2 = 9 times 2/6 = 3 A So change q = I times t = 34 times 5 = 15 C V_c = V_A = 12 V U_A = 0, V_C = 12 V [As +ve terminal is emf is connected to C] Potential at A = 0 V Potential at B = V_B- V_A = 2 I = V_B - 0 = 2 times 3 V_B = 6 V Potential at B = 6 v Point l = 12 V Power dissipated = VI = v^2/R = 12^2/2 = 72 W.