Question

4) In the circuit for problem 3, one of the resistors R2 is removed from the circuit so that the circuit just becomes R1, the other Rz, and the battery in series. Relative to the circuit in problem 3, the power dissipated in the remaining Ry will: (See answers on the other pages.) R W + 5) What is the total current through the battery for the circuit shown? Ra ww R3 w 루 W R2

Answer 1

* Answer -> (4) one of the desistors Rg is removed from the circuit which is parallel to Other Ry resistor so, initially our circuit Ener R2 R2 is like RI •1/2 1 I So, total current initial. I: - Req. where Reg = R + Rzeq [RI R, and Rzeq are in series) Rz• R₂ 'Both R2 in Parallel R2 + R2 : Rzeq = R2 Q Where. Rzeqa CB ingen .: Req = R1 + R2 E Total Current is i (R+R). aiso, Current is equally deveded in both px so, current passes through both P2 is I

so, perticular power disipattted in R2 is Prz (IV) ². R2 I ² P2 (IST) 2 2 same as for Other R2. so, initialy power in R2 is IER Now, if we remove one Pz from Circuit tuan Circzeit become muhi ]ع Ž R2 E Total Current in circuit Ig : Req Here Req = RitR2 [R, and Rs are in series E - (2) Rit R₂ s. If Here, we can able to compare (1) and (2) we brut. From that can say that I > If initiavy Current is more than in final condition in this situation Ff is freely passing through R2 also If > Il 4 If² R PRas . power disipattion is more in Re in final condition

so the final answer is power dissipated in the remaining R2 will * Increas, the current through P2 will increas in (final condition) (K) None of the above. Here, for all options (A) Battery current decrease clue to removal of R2 Cso, wrong) (B) Ry voltage dose not change. (C) The current through Rz is not exec they doubles. (it increas, but irrot doubles) CD) There will be achange we mension abore cles It will change (F) no matter what kind of combination (G) NOT decrease (Power dissipation) (H) NOT decrease (1) NOT decrease (K) Wone of above is sight answer.

(5) For a given Circreit wi Y UI Total Current 3 R I V Req R4 and Rg are Jo. equivalent of Now From the circuit In parallel connection both is R34 = Rso Ry Ret Ry R so, circuit is become like from circuit RR2 and R34 are in series - 834 " Req= R1 + R₂ + R34 R2 = R + R2 + R3.Ru (R3+ Ry? Req = Ro (Rgt R4) + R (R₂ + R4) + R3-Ry (R3 + Ry) Now, total £ is From Iz Y/Req 2. T - R, (R3784) +22(R3+R4) + R3 • Ry (P3 + Ry) I V (R3+ Ry) R(R3 +R;) + R (B3+R4) + R₂ oRy This, we can able to find total arent from the battery.