12th Physics || किरचॉफ के नियम Kirchhoff's Current Law || विद्युत धारा Ch 3 L-7 | By Gopal Sir

Revision|Electric Current (विद्युत धारा)|Resistance (प्रतिरोध) And Ohm's Law (ओम के नियम)|OMR

Find currents in different branches of th electric circuit shown in figure. How to Proceed : In this problem there are three wires EFAB, BE and BCDE . Therefore, we have three unknown currents i_1, i_2 and i_3 . So, we require three equations. One equation will be obtained by applying Kirchhoff's junction law (either at B or at E ) and the remainig two equations, we get from te second law (loop law). We can make three loops ABEFA, ACDF and BCDEB . But we have to chose any two of them. Initilly, we can choose any arbitrary directions of i_1, i_2 and i_3 .

Consider the circuit shown below. When switch S_(1) is closed, let I be the current at time t, then applying Kirchhoff's law E-iR-L (di)/(dt) = 0 or int_(0)^(i) (di)/(E-iR) = 1/L int_(0)^(1) dt i=E/R (1-e^(-R/l *t)) L/R = time constant of circuit When current reaches its steady value (=i_(0) , open S_(1) and close S_(2) , the current does reach to zero finally but decays expnentially. The decay equation is given as i=i_(0)e^(-R/L *T) ). When a coil carrying a steady current is short circuited, the current decreases in it eta times in time t_(0) . The time constant of the circuit is

Consider the circuit shown below. When switch S_(1) is closed, let I be the current at time t, then applying Kirchhoff's law E-iR-L (di)/(dt) = 0 or int_(0)^(i) (di)/(E-iR) = 1/L int_(0)^(1) dt i=E/R (1-e^(-R/l *t)) L/R = time constant of circuit When current reaches its steady value (=i_(0) , open S_(1) and close S_(2) , the current does reach to zero finally but decays expnentially. The decay equation is given as i=i_(0)e^(-R/L *T) . Three idenctical rings. The first (a). slips without rolling and the second (b) rolls without slipping and the third rolls with slipping .

Consider the circuit shown below. When switch S_(1) is closed, let I be the current at time t, then applying Kirchhoff's law E-iR-L (di)/(dt) = 0 or int_(0)^(i) (di)/(E-iR) = 1/L int_(0)^(1) dt i=E/R (1-e^(-R/l *t)) L/R = time constant of circuit When current reaches its steady value (=i_(0) , open S_(1) and close S_(2) , the current does reach to zero finally but decays expnentially. The decay equation is given as i=i_(0)e^(-R/L *T) . A wire is sliding as shown in fig . The angle between the acceleration and velocity of the wire is

This question is based on the following alphabet series. A B C D E F G H I J K L M N O P Q R S T U V W X Y Z If the first 5 letters are written in reversed order, then the next five letters are written in reversed order and so on, which is 7th letter to the right of 9th letter from the left? 1) T 2) S 3) P 4) Q 5) None of these

If the first half of English alphabet is reversed and the second half of the English alphabet,4s left undisturbed then answer the question given below. A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Which letter will be 7th to the left of 5th letter from the right? 1) O 2) X 3) L 4) B 5) None of these

If the first half of the English alphabet is reversed and then next portion of English alphabet is reversed so as ‘A’ takes the position of 'M’ and ‘N’ takes the position of 'Z' then answer the given question. A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Which letter will be 4th to the left of 7th letter to the right of 8th letter from the left? 1) B 2) C 3) K 4) L 5) None of these