Tiêu đề 3.CURRENT ELECTRICITY - Explanations.pdf ✅

1 (b) Maximum number of resistances = 2 n−1 = 2 3−1 = 4 2 (b) When wire is cut into two equal parts then power dissipated by each part is 2P1 So their parallel combination will dissipate power P2 = 2P1 + 2P1 = 4P1 Which gives P2 P1 = 4 3 (b) P = V 2 R ⇒ P1 P2 = R2 R1 ⇒ 6 P2 = 4 6 = 2 3 ⇒ P2 = 9 W 4 (c) Since H ∝ i 2 , so on doubling the current, the heat produced and hence the rise in temperature becomes four times 5 (a) According to Seebeck effect 6 (d) The resistance of an ideal voltmeter is considered as infinite 7 (b) R = ρl a for first wire and R ′ = ρl 4a = R 4 for second wire 8 (d) E = 2.2 V, R = 4 Ω, V = 2V r = ( E V − 1) R = ( 2.2 2 − 1) × 4 = 0.1 × 4 = 0.4Ω 10 (c) Initially the inductance will oppose the current which tries to flow through the inductance. But 10Ω and 20Ω can conduct. The current will be 2V 30Ω = 1 15 A 11 (a) 1 R1 = 1 10 + 1 2.5 = 5 10 = 1 2 ⇒ R1 = 2Ω Now 2Ω and 10 Ω are in series R2 = 10 + 2 = 12Ω R2 and 12Ω are in parallel 1 R3 = 1 12 + 1 12 ⇒ R3 = 6Ω Now R3and 6Ω are in series R4 = 10 + 6 = 16Ω Now, R4and 16 Ω are in parallel ∴ 1 R = 1 16 + 1 16 ⇒ R = 3Ω 12 (c) Energy = V 2 R t = 200 × 200 × 2 80 = 1000 Wh 14 (d) Thermocouples are widely used type of temperature sensor and are used as means to convert thermal potential difference into electric potential difference, using different combinations of metals. If metals were arranged according to their heating contrasts a series were formed antimony, iron, zinc, silver, gold, lead, mercury, copper, platinum and bismuth. The greater the heating contrasts between metals, the greater the electromotive force (EMF). Antimony and bismuth formed the best junction for emf. 15 (c) Potential difference across 1 MΩ resistor is VP − VB = 18V × 1 × 106Ω (0.2 + 1) × 106Ω = 18V × 1 × 106Ω 1.2 × 106Ω = 15V VB = −15V [Given] ∴ VP − VB = 15V or VP = 15V + VB = 15V − 15V = 0V Potential difference across 200 kΩ resistor is VA − VP = 18V × 0.2 × 106Ω (0.2 + 1) × 106Ω
= 18V × 0.2 × 106Ω 1.2 × 106Ω = 3V VA = +3V [Given] ∴ VA − VP = 3V or VP = VA − 3V = + 3V − 3 V = 0 V 16 (b) A = πr 2 = ρl⁄R or r = (ρl⁄πR) 1/2 r = ( 1.7 × 10−8 × 0.5 3.14 × 2 ) 1/2 = 0.367 mm 17 (b) According to Kirchhoff’s first law (5A)+(4A)+(-3A)+(-5A)+I=0 Or I=-1A 18 (b) I = 4 − 0.08t A Or dq dt = 4 − 0.08t A or q = ∫ (4 − 0.08t)dt C 50 0 or Ne = [4t − 0.08t 2 2 ] 0 50 = 100 C Where N is number of electrons. or N = 100 e = 100 1.6×10−19 = 6.25 × 1020 19 (c) Resistance of bulb is constant P = v 2 R ⇒ ∆P P = 2∆V V + ∆R R ∆P P = 2 × 2.5 + 0 = 5% 20 (a) E = xl = iρl ⇒ i = E ρl = 2.4 × 10−3 1.2 × 5 = 4 × 10−4A 21 (d) Let a resistance r ohm be shunted with resistance S, so that the bridge is balanced. If S’ is the resultant resistance of S and r, then In balanced position P Q = R S ′ 2 2 = 2 S ′ ∴ S ′ = 2Ω Now, 1 S ′ = 1 S + 1 r or 1 r = 1 S ′ − 1 S = 1 2 − 1 3 = 3 − 2 6 1 r = 1 6 r = 6Ω 22 (c) i = ev = 1.6 × 10−19 × 6.8 × 1015 = 1.1 × 10−3amp 23 (a) H1 = V 2 R t H2 = V 2 R/2 t ∴ H2 H1 = 2 ⇒ H2 = 2H1 24 (c) 4 A X 3 A 5 A 5 A l r S R P Q
Heat H = V 2 t R ⇒ H ∝ 1 R [If V,t constant] ⇒ HS HP = RP RS = ( R × 2R 3R ) (R + 2R) = 2 9 25 (c) In closed loop ABGFEHA 10 − i2 × 1 + i1 × 0.5 − 6 = 0 0.5i1 − i2 = −4 In closed loop BCDEB (i1 + i2 ) × 12 + i2 × 1 − 10 = 0 12ii + 13i2 = 10 From Eqs. (i) and (ii) i2 = 2.87A 26 (b) Heat produced by the heater H = V 2 R × t For 220V heater heat produced H1 = (220) 2 R × 5 For 110V heater heat produced H2 = (110) 2 R × L Now, H1 = H2 110 × 110 R t = 220 × 220 × 5 R t = 20 min 27 (c) I ∝ Q Ig I = 1 2 S G + S = 1 2 40 G + 40 = 1 2 ⟹ G = 40Ω 28 (c) An α −particle has a charge equal to 2 protons. Motion of α particle to the left, motion of proton towards left and motion of electrons towards right, all will produce conventional current towards left. The total current will be i = 1019 × (2 × 1.6 × 10−19) +1019 × (1.6 × 10−19) + 1019 × (1.6 × 10−19) = 6.4 A 29 (c) E = 2.2volt, V = 1.8 volt, R = 5R r = ( E V − 1) R = ( 2.2 1.8 − 1) × 5 = 1.1Ω 30 (b) The filament of the heater reaches its steady resistance when the heater reaches its steady temperature, which is much higher than the room temperature. The resistance at room temperature is thus much lower than the resistance at its steady state. When the heater is switched on, it draws a larger current than its steady state current. As the filament heats up, its resistance increases and current falls to steady state value 31 (c) We know that when current flow is same then resistors are connected in series, hence resultant resistance is R ′ = R1 + R2 = 10Ω + 20Ω = 30Ω Also since, cell are connected in opposite directions, the resultant emf is E = E1 − E2 = 5V − 2V = 3V From Ohm’s law E = iR ∴ i = E R = 3 30 = 0.1A 32 (a) vd = i⁄n Ae ; where n = Nρ/M = 6.023 × 1026 × 9 × 103 /63 = 0.860 × 1029 = 8.6 × 1028 and A = πD 2⁄4 = 22 7 × (10−3 ) 2 4m2 ⁄ ; = 11 14 × 10−6m2 vd = 1.1 8.6 × 1028 × 11 14 × 10−6 × 1.6 × 10−19 = 1 9.6 × 10+3 = 100 × 10−4 96 = 1.0 × 10−4m/s = 0.1 mms −1
33 (b) m = zit If V is the volume, then V= m ρ = zit ρ Thickness = zit Aρ = 3.3×10−7×1.5×20×60 50×10−4×9000 Thickness = 1.3 × 10−5m 34 (b) Let the voltage across any one cell is V, then V = E − ir = E − r1 ( 2E r1 + r2 + R ) But V = 0 ⇒ E − 2Er1 r1 + r2 + R = 0 ⇒ r1 + r2 + R = 2r1 ⇒ R = r1 − r2 35 (d) Current through arm ABC, = 4/(40 + 60) = 0.04 A Potential difference across A and B, VA − VB = 0.04 × 40 = 1.6V Current through arm ADC, = 4/(90 + 110) = 4/200A Potential difference between A and D, VA − VD = 4 200 × 90 = 1.8V ∴ VB − VD = (VA − VD) − (VA − VB) 1.8 − 1.6 = 0.2 V 36 (a) Voltage across each bulb V ′ = 110 2 = 55W so, power consumed by each bulb will be P ′ = ( 55 220) 2 × 500 = 125 4 W 37 (c) V 2 R + V 5 5 = 4 × ( V 2 R + 5 ) or R + 5 5R = 4 R + 5 On solving, we get R = 5Ω. 38 (b) m = Z i t ⇒ 20 × 10−3 = ( 32 96500) × 0.15 × t = 6.7 min = 6 min. 42 sec 39 (a) With rise in temperature the thermal velocity of the electron increases. Relaxation time and hence drift velocity will decrease. 40 (b) P1 = P2 = 60 W; when bulbs are connected in series then total power Ps = P1P2 P1+P2 = 60×60 60+60 =30 W 41 (c) Total resistance of the circuit = 80 2 + 20 = 60 Ω ⇒ Main current i = 2 60 = 1 30 A Combination of voltmeter and 80Ω resistance is connected in series with 20Ω, so current through 20Ω and this combination will be same = 1 30 A Since the resistance of voltmeter is also 80Ω, so this current is equally distributed in 80Ω resistance and voltmeter [i. e. 1 60 A through each] P.D. across 80Ω resistance = 1 60 × 80 = 1.33V 42 (d) i 2Rt = ms ∆t ⇒ (4) 2 × 7 × 3 × 60 r1 E r2 E V R 500W 55V 55V 110 V 500W 220V 220V