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Question: The current flowing through the segment \(AB\) of the circuit shown in figure is ![](https://ww...

The current flowing through the segment ABAB of the circuit shown in figure is

(A) 1  amp1\;{\rm{amp}} from AA to BB
(B) 1  amp1\;{\rm{amp}} from BB to AA
(C) 2  amp2\;{\rm{amp}} from AA to BB
(D) 2  amp2\;{\rm{amp}} from BB to AA

Explanation

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 AA and BB are joined using a metal wire. Hence, the potential difference across the points AA and BB is zero. Thus

Let us define R1=1  Ω{R_1} = 1\;\Omega , R2=3  Ω{R_2} = 3\;\Omega , R3=2  Ω{R_3} = 2\;\Omega and R4=4  Ω{R_4} = 4\;\Omega . From the circuit diagram, the resistances R1{R_1} and R2{R_2} are connected parallel to each other. Similarly, the resistances R3{R_3} and R4{R_4} are connected parallel to each other.

The equivalent resistance of the resistances R1{R_1} and R2{R_2} can be written as

R1,2=R1R2R1+R2{R_{1,2}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}

Here R1,2{R_{1,2}} is the equivalent resistance of the resistances R1{R_1} and R2{R_2}.

Substituting the values for R1{R_1} and R2{R_2} in the above equation, we get

R1,2=1  Ω×3  Ω1  Ω+3  Ω R1,2=34  Ω\begin{aligned} {R_{1,2}} &= \dfrac{{1\;\Omega \times 3\;\Omega }}{{1\;\Omega + 3\;\Omega }}\\\ \Rightarrow {R_{1,2}} &= \dfrac{3}{4}\;\Omega \end{aligned}

Again, we can write the equation for the equivalent resistance of the resistances R3{R_3} and R4{R_4} as

R3,4=R3R4R3+R4{R_{3,4}} = \dfrac{{{R_3}{R_4}}}{{{R_3} + {R_4}}}

Here R3,4{R_{3,4}} is the equivalent resistance of the resistances R3{R_3} and R4{R_4}.

Substituting the values for R3{R_3} and R4{R_4} in the above equation, we get

R3,4=2  Ω×4  Ω2  Ω+4  Ω R3,4=8  Ω26  Ω R3,4=43  Ω\begin{aligned} {R_{3,4}} &= \dfrac{{2\;\Omega \times 4\;\Omega }}{{2\;\Omega + 4\;\Omega }}\\\ \Rightarrow {R_{3,4}} &= \dfrac{{8\;{\Omega ^2}}}{{6\;\Omega }}\\\ \Rightarrow {R_{3,4}} &= \dfrac{4}{3}\;\Omega \end{aligned}

From the circuit diagram we can see that the combination of parallel resistances R1{R_1} and R2{R_2} is in series with the combination of parallel resistances R3{R_3} and R4{R_4}. Hence, we can write the equation for the equivalent resistance of the four resistances as,

R=R1,2+R3,4R = {R_{1,2}} + {R_{3,4}}

Here RR is the equivalent resistance of the four resistances.

Now, we can substitute 34  Ω\dfrac{3}{4}\;\Omega for R1,2{R_{1,2}} and 43  Ω\dfrac{4}{3}\;\Omega for R3,4{R_{3,4}} in the above equation to get,

R=34  Ω+43  Ω R=(3×3+4×44×3)  Ω R=2512    Ω\begin{aligned} R &= \dfrac{3}{4}\;\Omega + \dfrac{4}{3}\;\Omega \\\ \Rightarrow R &= \left( {\dfrac{{3 \times 3 + 4 \times 4}}{{4 \times 3}}} \right)\;\Omega \\\ \Rightarrow R &= \dfrac{{25}}{{12}}\;\;\Omega \end{aligned}

Now the total current flowing through the circuit can be written as

I=VRI = \dfrac{V}{R}

Here VV is the voltage across the circuit.

Since V=25  VV = 25\;{\rm{V}} and R=2512  ΩR = \dfrac{{25}}{{12}}\;\Omega , we can substitute the values for RR and VV to get the total current. Hence,

I=25  V2512  Ω I=12  A\begin{aligned} I &= \dfrac{{25\;{\rm{V}}}}{{\dfrac{{25}}{{12}}\;\Omega }}\\\ \Rightarrow I &= 12\;{\rm{A}} \end{aligned}

Now, using the current divider rule, we can write the equation for the current I1{I_1} flowing between point PP to AA as,

I1=R2R1+R2I{I_1} = \dfrac{{{R_2}}}{{{R_1} + {R_2}}}I

Substituting the values for R1{R_1}, R2{R_2} and II in the above equation, we get

I1=3  Ω1  Ω+3  Ω×12  A I1=34×12  A I1=9  A\begin{aligned} {I_1} &= \dfrac{{3\;\Omega }}{{1\;\Omega + 3\;\Omega }} \times 12\;{\rm{A}}\\\ \Rightarrow {I_1} &= \dfrac{3}{4} \times 12\;{\rm{A}}\\\ \Rightarrow {I_1} &= 9\;{\rm{A}} \end{aligned}

Again, using the current divider rule, we can write equation for the current I3{I_3} flowing from point BB to QQ as,

I3=R3R3+R4I{I_3} = \dfrac{{{R_3}}}{{{R_3} + {R_4}}}I

Substituting the values for R3{R_3}, R4{R_4} and II in the above equation, we get

I3=4  Ω2  Ω+4  Ω×12A I3=46×12A I3=8  A\begin{aligned} {I_3} &= \dfrac{{4\;\Omega }}{{2\;\Omega + 4\;\Omega }} \times 12\,{\rm{A}}\\\ \Rightarrow {I_3} &= \dfrac{4}{6} \times 12\,{\rm{A}}\\\ \Rightarrow {I_3} &= 8\;{\rm{A}} \end{aligned}

Now, the current flowing through the segment ABAB can be written as

i=I1I3i = {I_1} - {I_3}

Here ii is the current flowing through the segment ABAB.

Now substituting the values of I1{I_1} and I3{I_3}, we get

i=9  A8  A i=1  A\begin{aligned} i &= 9\;{\rm{A - 8}}\;{\rm{A}}\\\ \Rightarrow i &= 1\;{\rm{A}} \end{aligned}

Since, the current obtained has a positive value, we can confirm that the direction of the current is from AA to BB.

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.