Question
Question: The current flowing through the segment \(AB\) of the circuit shown in figure is  1amp from A to B
(B) 1amp from B to A
(C) 2amp from A to B
(D) 2amp from B to A
Solution
First, we can redraw the circuit diagram so that it is easy to identify the parallel resistances and the resistances in series with each other. We can apply the current divider rule to the circuit to find the solution.
Complete step by step answer: Let us first redraw the given circuit diagram as follows:
Here the points A and B are joined using a metal wire. Hence, the potential difference across the points A and B is zero. Thus
Let us define R1=1Ω, R2=3Ω, R3=2Ω and R4=4Ω. From the circuit diagram, the resistances R1 and R2 are connected parallel to each other. Similarly, the resistances R3 and R4 are connected parallel to each other.
The equivalent resistance of the resistances R1 and R2 can be written as
R1,2=R1+R2R1R2
Here R1,2 is the equivalent resistance of the resistances R1 and R2.
Substituting the values for R1 and R2 in the above equation, we get
R1,2 ⇒R1,2=1Ω+3Ω1Ω×3Ω=43ΩAgain, we can write the equation for the equivalent resistance of the resistances R3 and R4 as
R3,4=R3+R4R3R4
Here R3,4 is the equivalent resistance of the resistances R3 and R4.
Substituting the values for R3 and R4 in the above equation, we get
R3,4 ⇒R3,4 ⇒R3,4=2Ω+4Ω2Ω×4Ω=6Ω8Ω2=34ΩFrom the circuit diagram we can see that the combination of parallel resistances R1 and R2 is in series with the combination of parallel resistances R3 and R4. Hence, we can write the equation for the equivalent resistance of the four resistances as,
R=R1,2+R3,4
Here R is the equivalent resistance of the four resistances.
Now, we can substitute 43Ω for R1,2 and 34Ω for R3,4 in the above equation to get,
R ⇒R ⇒R=43Ω+34Ω=(4×33×3+4×4)Ω=1225ΩNow the total current flowing through the circuit can be written as
I=RV
Here V is the voltage across the circuit.
Since V=25V and R=1225Ω, we can substitute the values for R and V to get the total current. Hence,
I ⇒I=1225Ω25V=12ANow, using the current divider rule, we can write the equation for the current I1 flowing between point P to A as,
I1=R1+R2R2I
Substituting the values for R1, R2 and I in the above equation, we get
I1 ⇒I1 ⇒I1=1Ω+3Ω3Ω×12A=43×12A=9AAgain, using the current divider rule, we can write equation for the current I3 flowing from point B to Q as,
I3=R3+R4R3I
Substituting the values for R3, R4 and I in the above equation, we get
I3 ⇒I3 ⇒I3=2Ω+4Ω4Ω×12A=64×12A=8ANow, the current flowing through the segment AB can be written as
i=I1−I3
Here i is the current flowing through the segment AB.
Now substituting the values of I1 and I3, we get
i ⇒i=9A−8A=1ASince, the current obtained has a positive value, we can confirm that the direction of the current is from A to B.
Therefore, the option (A) is correct.
Note: It should be noted that the resistances connected between two pints at the same potential difference are in parallel. Similarly, we should also note that the resistances connected between points at different potential differences are in series.