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In the circuit shown in Figure E=12V,C_(1)=4 muF,C_(2)=2muF,C_(3)=6 muF and C_(4)=3muF Find the heat produced in the circuit after switch S is shorted.

A particle travels m a straight line, such that for a short time 2 s le tle 6 s, its motion is described by v= (4/a) m//s, where a is in m//s^2. If v= 6 m//s. when t= 2 s, determine the particle's acceleration when t= 3 s.

When an ammeter of negligible internal resistance is inserted in series with circuit, it reads 1 A . When a voltmeter of very large resistance is connected across R_(1) , it reads 3 V . But when the points A and B are short-circuited by a conducting wire, then the voltmetre measures 10.5 V across the battery. The internal resistance of the battery is equal to

If voltage applied to X -ray tube increased from V=10kV to 20kV . The wave length interval between K_(alpha)- line and short wave cut off continuous X -ray increases by factor of 3 (Rhc)/e=13.6V where R is Rydberg constant h is plank's constant c is speed of light in vacuum and e is charge on electron

In the figure shown V_(1),V_(2),V_(3) are AC voltmeters and A is AC ammeter. The readings of V_(1), V_(2), V_(3) and 10 V, 20 V, 20 V, 2A respectively. If the inductor is short circuited, then

In the circuit in figure E_1=3V, E_2=2V, E_3=1V and R=r_1-r_2-r_3=1Omega a. Find the potential differece between the points A and B and the currents through each branch. b. If r_2 is short circuited and the point A is connected to point B , find the currents through E_1,E_2, E_3 and the resistor R

Rectify the following errors: (i) Purchases Book is overcast by Rs. 500. (ii) Salary paid to an employee, Mr. Ajay, is debited to his Personal Account Rs. 3,000. (iii) Goods sold to Shashi on credit Rs. 300 have been wrongly passed through the Purchases Book. Total of Returns Inward Book has been added Rs. 9 short. (v) Purchase of chair from Happy Traders for Rs. 35 has been entered in the Purchases Book as Rs. 53.

A capacitor of capacity 1muF is connected in a closed series circuit with a resistance or 10^(7) ohms, an open key and a cell of 2 V with negligible internal resistance : (i) When the key is switched on at time t=0, find, (a) The time constant for the circuit. (b) The charge on the capacitor at steady state. (c) Time taken to deposit charge equal to half of charge that will deposit at steady state. (ii) If after completely charging the capacitor, the cell is shorted by zero resistance at time t=0, find the charge on the capacitor at t=50 s. (Given : e^(-5)=6.73xx10^(-3), ln^(2)=0.693 )