Question Calculate the electric current in the given circuit when
(a) Key K1 is open and K2 is closed
(b) both keys are closed
(c) Key K2 is open and K1 is closed
Solution .
(a) When, key K1 is open and K2 is closed , then no current flows in the circuit as circuit is an open circuit.
(b) When both keys are closed a current begin to flow in the circuit. Let us consider the circuit given below where we have labelled above given circuit.
How to determine equivalent resistance:-
(i) Now if you look closely at the circuit you would find that current is dividing at point A and combining again at point B.
(ii) Same amount of current (say I1) will flow through resistors R1 and R2 also same amount of current (say I2) will flow through resistors R3 and R4.
(iii)We are aware of the fact that “For a series combination of resistors the current is same in every part of the circuit or same current flows through each resistor”. So we can say that resistors R1 and R2 are connected in series combination. Similarly resistors R3 and R4 are also connected in series combination.
(iv) again as mentioned above in step (i) current is dividing at point A so we have different currents flowing through the combination of resistors R1 and R2 and resistors R3 and R4.
(v)So equivalent resistance of resistors R1 and R2 which is (R1 + R2) and equivalent resistance of resistors R3 and R4 which is (R3 + R4) are connected in parallel combination to each other as they have different amount of current flowing through them.
(vi)So, Equivalent resistance in the circuit would be
Putting the values as given in the question we get
${\frac{1}{R} = \frac{1}{{4 + 4}} + \frac{1}{{4 + 4}} = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}}$
So,
$R=4\Omega$
Electric current,
$I = \frac{V}{R} = \frac{{12}}{4} = 3A$
(c) When key K2 is open and K1 is closed, the part ADB will become an open circuit, So no current will flow in this part of the circuit.
Therefore, net resistance of the circuit will be
$R = {R_1} + {R_2} = 4 + 4 = 8\Omega $
Therefore, electric current.
$I = \frac{V}{R} = \frac{{12}}{8} = 1.5A$
Watch this tutorial for learning about how to solve resistance problems for class 10.
Given below are the links of some of the reference books for class 10 Science.
You can use above books for extra knowledge and practicing different questions.
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