Direction: Q.41 to Q.45 |
In 1942, a German physicist Kirchhoffs extended Ohm's law to complicated circuits and gave two laws, which enable us to determine current in any part of such a circuit. According to Kirchhoffs first rule, the algebraic sum of the currents meeting at a junction in a closed electric circuit is zero. The current flowing in a conductor towards the junction is taken as positive and the current flowing away from the junction is taken as negative. According to Kirchhoffs second rule, in a closed loop, the algebraic sum of the emfs and algebraic sum of the products of current and resistance in the various emf of the loop is zero. While traversing a loop, if negative pole of the cell is encountered first, then its emf is negative, otherwise positive. |
Read the above passage carefully and give the answer of the following questions: |
A) 4.5 A
B) 3.7 A
C) 2.0 A
D) 2.5 A
Correct Answer: C
Solution :
(c) 2.0 A According to Kirchhoff's junction law \[(+l)+(+4\,A)+(+2\,A)+(-5\,A)+(-3\,A)=0\] \[l+6\,A-8\,A=0\] or \[l=2\,A\]You need to login to perform this action.
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