Tiêu đề 11-dc-circuit-and-capacitor-.pdf ✅

DC CIRCUIT AND CAPACITOR 1. (B) P = V 2 V2 P1 + V2 P2 + V2 P3 +⋯. 2. (B) Equivalent circuit. R1 R2 = R4 R3 3. (D) (1680 + r)I = 20 ⇒ (2930 + r)I = 30 ⇒ 2 × 2930 + 2r = 3 × 1680 + 3r ⇒ r = 820, I = 8mA 4. (D) Resistors whose resistances are written in brackets are parallel. 5. (A) E = jρ [j = current density ] j = I πr 2 [r = radius of cross section at distance ' x ' from left end ] r = [a + (b−a) l x] Hence, E = Vl 2ρ πR(al+(b−a)x) 2 6. (A) Resistance of each part = R 2n For 'n' such parts connected in series, equivalent resistance, say R1 = n [ R 2n ] = R 2 . Similarly, equivalent resistance, say R2 for another set of n identical resistors in parallel would be 1 n ( R 2n ) = R 2n2 For getting maximum of R1&R2, they should be connected in series & hence, Req = R1 + R2 = R 2 (1 + 1 n2 ) 7. (A) 5 × 10−3A = I(0.1Ω) (0.2+0.3+0.5) ⇒ I = 50 mA 8. (C) 3 − ( 18 3+R ) ⋅ 1 = 0 ⇒ R = 3Ω 9. (C) VA − VB = voltage drop across capacitor + voltage drop across resistor ∴ − 11 = Q C + iR ⇒ −11 = 16 4 + i. 3 × 103 ⇒ i = −5mA Power delivered by capacitor P = iV = (5mA)(4V) = 20mW 10. (D) Initial charging rate = initial current in the line of capacitor = 2E 5R Steady state p.d. across capacitor : V0 = 2 3 E ⇒ q0 = CV0 = 2 3 EC ⇒ t = q0 i = 2/3EC 2E/5R = 5 3 RC 11. (A) Suppose that the inner sphere is at a higher potential than outer sphere. Let the current be i. Consider a thin shell of thickness dr at a distance r from the centre. Let voltage across it be dV. Then, applying V = iR For thin shell: dV = i ⋅ dr σ4πr 2 ⇒ Edr = i ⋅ dr 4πr 2 ⋅ 1 kE ∴ E = √ i 4πkr 2 = C r where C = √ i 4πk
Using, E = − dV dr C r = − dV dr ⇒ C(lnr)a b = Va − Vb V = Va − Vb = Cln (b/a) V = √ i 4πk ln (b/a) ⇒ i = V 24πk [ln p(b/a)] 2 12. (D) ⇒ P.d. across C is zero ⇒ charge = 0 13. (D) Let n1: no. of capacitors be connected in parallel, n2: no. of such parallel combination connected in series. n2 = 1000 250 = 4 and n1(8μF) n2 = 16μF ⇒ n1 = 8 ⇒ Total no. of capacitors required = 32 14. (B) Charge on C = sum of charges at 4&5 = 2CV = 2 × 2 × 10 = 40μC 15. (B) ⇒ inside dielectric, field = σ εε0 = σ−σ ′ ε0 ⇒ σ ′ = ( ε−1 ε ) σ ⇒ Force between A&M = ( ε−1 ε ) σA ⋅ σ 2ε0 (Using Eq = F ) = σ 2 (ε − 1)A 2 ∈ ε0 16. (B) Current is obviously constant, by charge conservation. Using i = n0eAvd, we can say that if A is non-uniform, vd will be non-uniform. Similarly, since vd depends on electric field, electric field will also be non uniform. 17. (AC) Current is obviously constant, by charge conservation. Using i = n0eAvd, we can say that if A is non-uniform, vd will be inversely proportional to A. Similarly, since vd increases with electric field, electric field at A will be more than B. 18. (A) Mark the voltages as shown in figure: By KCL, sum of all currents emerging from point P is zero i.e. x−8 2 + x−6 5 + x−0 4 + x−4 3 = 0. Solving this, we can get x. 19. (ABC) Let the point where jockey touches wire be called S. Then the direction of current shown in figure indicates that voltage across QS is less that E2. This can happen if: 1. E1 is too low 2. r is too high ( if r is too high, it will take up more voltage and less will be left for QS). 20. (BD) Let the currents be as shown in the figure: KVL along ABCDA ⇒ −10i − 2 + (2 − i)1 = 0 ∴ i = 0
Potential difference across S = (2 − i)1 = 2 × 1 = 2 V. 21. (ABCD) After redrawing the circuit (a) I4 = 5 A (b) From loop (1) −8(3) + E1 − 4(3) = 0 ⇒ E1 = 36 volt (c) From loop (2) + 4(5) + 5(2) − E2 + 8(3) = 0 E2 = 54 volt (d) From loop (3) −2R − E1 + E2 = 0 R = E2 − E1 2 = 54 − 36 2 = 9Ω 22. (AC) Equivalent diagram is as shown. If P is moved 2 cm right, then R1 = 12Ω, R3 = 3Ω R1 R3 = R2 R4 (Hence wheat stone will be balanced.) If S is moved left by 5 3 cm, then R3 = 10 3 and R4 = 20 3 hence R1 R3 = R2 R4 (hence wheatstone's bridge will be balanced.) 23. (AB) R AC = R1 + R2 CB ⇒ R L/4 = R1 + R2 3L/4 ⇒ R1 + R2 = 3R and R + R1 2l/3 = R2 l/3 ⇒ R + R1 = 2R2 R2 = 4R 3 and R1 = 5R 3 24. (AC) 10 3 − 5 2 1 3 + 1 2 = εeq 5/6 5/6 = 1 V req = 6 5 ⇒ i = 1 4 5 + 6 5 = 0.5 A P.d. = 0.5 × 4 5 = 0.4 V ; 0.5 × 4 5 = 0.4 V −3i1 + 10 − 1.0 − 0.4 = 0 ⇒ i1 = +ve −0.4 − 2i2 − 5 = 0 ⇒ i2 = −ve 25. (AB) If P is opened effective resistance would increase ∴ A is correct & C is wrong. As drop across R1 is reduced ∴ B is correct. Battery is ideal hence D is wrong. 26. (AB) The effective emf of secondary is 0 V ∴ A& B are correct and D is wrong. There will be current in 1 V battery ∴ C is wrong. 27. (ABC) ∵ B2 and B3 are in series, ∴ i2 = i3; Also, V2 + V3 = V1 ∵ P1 = P2 = P3 = P4
⇒ P2 = P3 or i2 2R2 = i2 2R3 ⇒ R2 = R3 ∴ V2 = V3 = V1 2 ⇒ V1 2 R1 = V1 2 R1 = V1 2 36 and P3 = V3 2 R3 = V1 2 4R3 V1 2 36 = V 2 4R3 ⇒ R3 = 9Ω = R P1 = P4 ⇒ I1 2R1 = I4 2R4 or ( I4 3 ) 2 × 36 = I4 2R4 ⇒ R4 = RΩ Voltage output of battery = V1 + V4 [P1 = 4 = V1 2 36 ⇒ V1 = 12 V&P2 = 4 = V4 2 4 ⇒ V4 = 4 V] ∴ V1 + V4 = 16 V 28. (ABCD) i1 = VA−VB 6 = 10 6 = 5 3 A i2 = i1 + 1 = 5 3 + 1 = 8 3 A VE − VD = VA − VB = 10 ⇒ i3 = 10 3 A i1 + i3 = 5 3 + 10 3 = 5A in loop FEDC ε2 − 3i3 − 4(i1 + i3 ) − 3(2 + i1 + i3 ) = 0 ⇒ ε2 = 51 V in loop FGHC ε1 − 4 × 2 − 3(2 + i1 + i3 ) = 0 ⇒ ε1 = 29 VR = VA − VG 1 = ε2 − ε1 1 = 22Ω 29. (ABCD) (1) 4x − 10y − 6(I − x) = 0 (2) 3(x + y) − 7(I − X − y) + 10y = 0 (3) 6(I −x)+ 7(I −x −y)+2I − 49 = 0 x = 259 52 A = 4.98A y = 4.98 × 5 74 = 0.34A I = 5(4.98 −0.34) 3 = 7.73 A I − x − y = 2.41A I − x = 7.73 − 4.98 = 2.75 30. (ACD) At t = 0, capacitor is uncharged and there will be no voltage across it. Hence it can be short- circuited in solving the circuit. After a long time, no current flows through the branch containing capacitor. i.e. rest of the circuit elements will be in series. 31. (ACD) At steady state : I(3) + I(2) = 15 I = 3 KVL C → D → E → a → b → C (V/C) − I(3) + q 11 − 7 + q 5 = 0 ⇒ q 11 + q 5 = 7 + 3 × 3 = 16 KVL: a → b ⇒ Va − 7 + q 5 = Vb ⇒ Va − Vb = 7 − q 5 = 7 − 55 5 = −4V P.d. across C1 ⇒ q 11 = 55 11 = 5 V; P.d. across C2 ⇒ q 5 = 11 V P.d. across terminal = 15 − I(2) = 15 − 3 × 2 = 9 V 32. (AB) We know that the capacitance of an empty capacitor increases k times if a dielectric is inserted in it. Therefore, in this case, the capacitance of combination will increase upon insertion of a dielectric. Also, by Q = CV, charge supplied by battery also proportionately increases for keeping V constant. 33. (ABC) At t = 0, capacitor will have no voltage across it. Hence A. Voltage across capacitor will gradually increase with time. Hence B. C can be calculated by the usual charging equation for capacitor. 34. (ACD) Q = CE Wbattery = Uf − Ui + ΔH E ⋅ CE = 1 2 CE 2 + ΔH ΔH = H1 = CE 2 − 1 2 CE 2 = 1 2 CE 2