Tiêu đề CURRENT ELECTRICITY (Passage Based).pdf ✅

PHYSICS Passage # 1 (Ques. 1 to 3) Each component in the infinite network shown in Fig. has a resistance R = 4 . A battery of emf 1V and negligible internal resistance is connected between any two neighbouring points, say X and Y. R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R R A 1V X Y Ideal ammeter Answer the following questions Q.1 Current shown by the ammeter will be - (A) 0. 25 A (B) 0.5 A (C) 0. 75 A (D) 1 A Sol. [B] We can consider the network to consist of two resistances connected in parallel between X and Y. One of these is the resistance R between X and Y and the other is the equivalent resistance of the rest of circuit. This is shown in Fig (A). R R R R R R R R R R R R R R R R R I/4 I/4 I/4 X I/4 I R R R R R R R R R R R R R R R R I/4 Y I/4 I/4 I/4 I R R R R R R R R R R R R R R R R Y R V a b I A I X I/4+I/4 =I/2 R'eq R V X Y I/2 I/2 I I (A) A (Here, R'eq is the equivalent resistance of rest of the circuit, i.e., excluding R) A Req X Y V I I (Here, Req is the equivalent of R'eq and R so Req is the equivalent resistance of total circuit) Referring to Fig (A), V = 2 I R (also V = I/2 R ́eq) and form fig (B), V = I Req So I Req = 2 I R
or Req = 2 R Hence the equivalent resistance of the network between X and Y or any two neighbouring points is R/2. In Fig. (B) V = I Req or I = Req V but Req = R/2  I = R 2V Given V = 1V, R = 4  I = 4 2 = 0.5 A So the correct answer is (B). Q.2 If the resistance R between X and Y is removed, current shown by ammeter will be - (A) 0.25 A (B) 0.5 A (C) 0.75 A (D) 1 A Sol. [A] A R'eq=R X Y V In Fig (A), since lthe current I is equally shared by R and R'eq, so R'eq = R. Now if the resistance R is removed, it will be only R'eq = R placed across the battery so that current will now be I = R V = 4 1 = 0.25 A (In this case equivalent resistance of circuit will be R'eq = R) Hence the correct option is (A). Q.3 Equivalent capacitance, if the given network consists of capacitors each C (instead of resistances), and with none of the components removed, will be - (A) C/4 (B) C/2 (C) 2C (D) 4C Sol. [C] Let us draw the corresponding figures for capacitances. C C C C Q Q/4 Q/4 Q/4 Q/4 X C C C C Q Q/4 Q/4 Q/4 Q/4 Y (A) (B) C C C Q X C C C C Q Y (C) Q/4+Q/4=Q/2 C'eq C V Y X Q/2 Q/2 Q Q (D) (C ́eq is the equivalent capacitance of rest of the circuit, i.e., excluding C) Ceq V Y (E) Q X (Ceq is the equivalent capacitance of total circuit between X and Y) In Fig (D) Potential difference across C, V = C Q/ 2 = 2C Q
in Fig (E) V = Ceq Q  Ceq Q = 2C Q or Ceq = 2C so the correct option is (C) Passage # 2 (Ques. 4 to 6) A fan operates at 200 volt (DC) consuming 1000 W when running at full speed. It's internal wiring has resistance 1 When the fan runs at full speed, its speed becomes constant. This is because the torque due to magnetic field inside the fan is balanced by the torque due to air resistance on the blades of the fan and torque due to friction between the fixed part and the shaft of the fan. The electrical power going into the fan is spent (i) in the internal resistance as heat , call it P1. (ii) in doing work against internal friction and air resistance producing heat, sound etc, call it P2. When the coil of fan rotates, an emf is also induced in the coil. This opposes the external emf applied to send the current into the fan. This emf is called back-emf, call it 'e'. Answer the following questions when the fan is running at full speed. Sol.4 [C], Sol.5 [C], Sol.6 [B] The fan is operating at 200 V, consuming 1000 W, then I = 200 1000 = 5A But as coil resistance is 1 then power dissipated by internal resistance heat is P1 = I2R = 25 W if V is net emf across coil then R V 2 = 25 W V = 5 volt Net emf = source emf – back emf V = Vs – e  e = 195 C The work done P2 = 1000 – 25 = 975 W Q.4 The current flowing into the fan and the value of back emf 'e' is - (A) 200 A, 5 Volt (B) 5 A, 200 Volt (C) 5 A, 195 Volt (D) 1A, 0 volt [C] Q.5 The value of power 'P1' is - (A) 1000 W (B) 975 W (C) 25 W (D) 200 W [C] Q.6 The value of power 'P2' is - (A) 1000W (B) 975 W (C) 25 W (D) 200 W [B] Passage # 3 (Ques. 7 to 9) R 2R R R 4R R 2R R R R A R 2R R 3R E r B 2 4 6 8 10 5 4 3 2 1 P (watt) R Cell emf is fixed and internal resistance r is also fixed. As value of R changes power delivered to circuit by cell also changes. Variation of power delivered with R is shown in graph. 7.[C], 8.[B], 9.[B] R/2 B R R 2R A R R 3R r R 3R 0.5R 3R r 1.5R 1.5R 1 1.5R 1 0.5R 1 R 1 eq = + + 1.5R 3 1 1 R 1 eq + + =  Req = 0.3R Power is maximum at R = 6
for maximum delivered power Rexteq. = r 0.3R = r 0.3 × 6 = r r = 1.8 r 0.3 R E i = 0.3R r E + = 2r E Pmax. = i2 × 0.3R = 2 2 4r E × r = 4r E 2 5 = 4 1.8 E 2  E = 6 volt current through battery = 2 1.8 6 2r E 0.3R r E  = = + p.d. across battery = 1.8 3 6− × 1.8 = 3V current through 3R = 6 1 R 1 3R 3 = = = 0.17A p.d. accross 2R = 3 2 × 3 = 2 volt Q.7 Emf of the cell is (A) 9V (B) 11.5 V (C) 6V (D) 4.5 V Q.8 Current through the resistance 3R is nearly (when R = 6) (A) 0.37 A (B) 0.17 A (C) 0.49 A (D) 0.12 A Q.9 Potential difference across each resistance 2R i.e. between A & B is nearly (when R = 6) (A) 3.5 V (B) 2V (C) 1.6 V (D) 1.2 V Passage 4 (Question 10 to 15) In the laboratory, the voltage across a particular circuit element can be measured by a voltmeter. A voltmeter has a very high resistance and should be connected in parallel to the circuit element whose voltage is being measured. Connected improperly, the voltmeter will affect the circuit, interrupting it and preventing current from flowing through the circuit element that it is meant to measure. An experiment is conducted in which a voltmeter is used to investigate voltages in a circuit containing a capacitor and a light bulb. The bulb and the capacitor are connected in series with a battery and the voltmeter is placed in different positions : in the first case across the capacitor, in the second case across the light bulb, and in the third case across the battery (see figure 1). The voltmeter reading is recorded every 10 seconds. The voltage for Case 1 as a function of time is shown in figure 2. Case 1 voltmeter V capacitor battery bulb Case 2 voltmeter V capacitor bulb battery Case 3 voltmeter V capacitor bulb battery Figure – 1 voltage time Figure – 2 A capacitor consists of two conducting plates separated by a non-conducting material. When a battery is connected to a circuit containing a capacitor and a light bulb in series, a current will flow, causing positive charge to accumulate on one capacitor plate and an equal amount of negative charge to accumulate on the other. After the current has flowed for a finite time, the capacitor will be fully charged. The ratio of the