Current Electricity-09-OBJECTIVE & SUBJECTIVE LEVEL I

Current Electricity LEVEL I OBJECTIVE (Solutions) 1. where is drift velocity of electron 2. Power Since they are in series grouping Power 3. 4. (where m is mass and D is density) and 5. Equivalent diagram is shown below 6. Let with a emf of to batteries with internal resistance and respectively then equivalent emf = may be a smaller and greater than individual emfs. But equivalent resistance = Which is less then 7. These three resistance are in parallel grouping 8. Using Kirchhoff’s Law (KVL) in loop PQRS …(i) and KVL for loop PQBD …(ii) solving (i) and (ii) we get 9. KVL for lop containing and R ….(i) KVL for loop containing and R …(ii) Solving (i) and (ii) and 10. In steady state there is no current / through capacitor and therefore 11. Solving 12. because at , 13. For zero deflection …(i) Current in upper branch = = p.d. per unit length = 14. 15. Since voltmeter is ideal no current will be drawn from it. Therefore equivalent resistance and current supplied by the battery Using KVL in loop abcda Level II (objective) 1. 2. 3. Each branch of the cube is assumed to be a resistor of resistance ‘R’ 4. Voltmeter is constructed using high resistance in series with galvanometer. 5. Let resistance R is added in series with galvanometer p.d through galvanometer …(i) p.d. across resistance = …(ii) equating (i) and (ii) 6. 7. 8. Power dissipated in R for P to be maximum amp 2nd method Both batteries are in parallel From maximum power theorem 9. 10. When switch is open no current will pass through OS. When switch is closed current through OS 11. Taking loop i.e. there is no current in any branch 12. Using Kirchhoff’s law in loop XYMLX …(i) Using Kirchhoff’s law in loop YZNMY (using eqn. (i)) 13. Le at any time ‘t’ and be the charge on ‘2C’ and ‘C’ respectively. Applying Kirchhoff’s law in loop containing 2C, R and C (upper half) …(i) KVL (lower loop) …(ii) Solving eq. (ii) Integrating Putting in eq. (i) increasing with time 14. When switch is closed for circuit Containing C, R and E time constant is RC i.e. when switch is opened both resistors are in series Time constant 15. …(i) …(ii) solving (i) and (ii) we get CBSE level – I 1. also 2. (a) Req = …(i) (b) 3. In steady state there is no current through capacitor branch. Current = 3 amp Current distribution is as shown below p.d. across capacitor 4. Maximum current that can be drawn 5. (a) Using balanced bridge condition (b) solving for from ‘A’ (c) Since battery is connected the balance bridge condition. Circuit will not drawn any current. Therefore deflection in galvanometer will be zero. 6. Let an external resistance R is added in series with galvanometer 7. (using ) 8. (where is drift velocity) time taken 9. Since potential at is zero and Current in the branch = =- 1 amp 10. (a) Terminal voltage = (b) 11. (a) (i) For maximum resistance they must be added in series (ii) For minimum resistance they must be added in parallel Ratio (b) (i) 1 and 2 are in parallel and resultant in series with 3 (ii) 2 and 3 as in parallel and resultant in series with 1  (iii) All are in series (iv) All are in parallel (c) (i) (ii) All resistances are in series 12. =2.73 13. In first case voltmeter has highest resistance it will draw minimum current and therefore it will give maximum reading. In third case (c) resistance of voltmeter is minimum, therefore it will give minimum reading 14. (discharging) …(i) differentiating with respect to time …(ii) from equation (i) and (ii) reading will be same at all times. 15.(a) Let resistance of rod be and external resistance be ‘r’ volt …(i) also …(ii) Using KVL (b) Subjective Level-II 1. Using KVL in loop acdbpqa …(i) Using KVL in loop acdfa …(ii) Using eqn. (i) and (ii) Total current supplied by battery 2. Using KVL in loop AEBDFA …(i) KVL in loop BDGCHB …(ii) multiplying eqn. (i) by 3 and eqn. (ii) by 1, we get (a) (b) Through G through H, 3. 4.(a) (b) 5. Using KVL in loop abgha KVL in loop bcfgb similarly 6. Total current supplied by battery applying Kirchhoff’s law in loop abcdefa power dissipated 7. After the closing the switch for very long time current through capacitor branch will become zero. current As per question p.d. across one the battery is zero. 8. Let external resistor be and internal resistance of battery then …(i) and …(ii) and …(iii) also …(iv) solving eqn. (ii), (iii) and (iv) 9. power …(i) as per question taking logarithm No of time constant 10. Total current supplied by the battery Amp Let Then Now Subjective level – III 1. (a) (b) Total current supplied by the battery I = . Current in branch BC is amp = 20 Volt (c) No, because current through BC wire still be amp. 2. (a) Total current (b) Let current in 36 resistor is therefore current in 18 is and is 12 (c) 3.(a) Let potential of point A is and that of B is zero (b) and Current (c) Effective emf = 2 volt effective resistance = 4. Ammeter will have minimum resistance while voltmeter will have maximum resistance. Note : Ammeter is always added in series with circuit while voltmeter is added in parallel with the branch. 5. In steady state current potential difference across resistor = for discharging = = 11 m amp. 6. Capacitance electric field between capacitor plate Resistance Potential difference V = Ed = Leakage current 7. Using Young’s modulus …(i) (change in resistance) = 8.(a) In steady state there is no current is capacitor branch = 4 Current in 2 resistance (b) Total resistance I = Reading of voltmeter 9.(a) Considering a sphere of radius ‘r’ and thickness ‘dr’ dR (small resistance) (b) Current J (Current density) = = 10. Initial charge on capacitor …(i) when switch S1 is closed the capacitor will start to discharge = …(ii) i.e. after time charge on capacitor = Next time (from to )