What is the equivalent resistance between the points A and B of the circuit shown in figure.
What is the equivalent resistance between the points A and B of the circuit shown in figure.
The correct option is A 7Ω
This is clearly not a Wheatstone bridge as R1R2≠R3R4
Which in turn implies that points C and D are not at the same potential and there will be a current through the 5Ω resistor.
Let’s simplify this and connect a battery of V volts across A and B. Also let’s say a total current of i is provided by the battery. Refer the given diagram below:
SYMMETRY ARGUMENTS:
If I modify the circuit, so that, the point B is at the place of A now and vice-versa. Would you be able to differentiate? No right!
This is called point symmetry. It also implies that the current distribution should also be symmetric, which simply means this-
If you are still not clear, you can see from the above diagram that the distribution of current remains the same while leaving the branches in the same order as they entered. This means that there will be a current of i1−i2 through CD as shown below:
Now the question is simplified. We can use Kirchhoff’s law to calculate i1 and i2 in terms of V and apply V=iReq to get Req.
Let’s do that. Refer the following diagram:
Loop ACDFA
−10i1−5(i1−i2)+5i2=0−15i1+10 i2=0i1=23i2
Loop ACEBGHA:
−10i1−5 i2+V=010 i1+5i2=V
Solving both the equations in finding i1 and i2 in terms of V, we get
i2=335 V
i1=235 V
i=i1+i2=535 V=17 V
Which implies, Vi=7Ω
Thus equivalent resistance between A and B is 7Ω.
7Ω
This is clearly not a Wheatstone bridge as R1R2≠R3R4
Which in turn implies that points C and D are not at the same potential and there will be a current through the 5Ω resistor.
Let’s simplify this and connect a battery of V volts across A and B. Also let’s say a total current of i is provided by the battery. Refer the given diagram below:
SYMMETRY ARGUMENTS:
If I modify the circuit, so that, the point B is at the place of A now and vice-versa. Would you be able to differentiate? No right!
This is called point symmetry. It also implies that the current distribution should also be symmetric, which simply means this-
If you are still not clear, you can see from the above diagram that the distribution of current remains the same while leaving the branches in the same order as they entered. This means that there will be a current of i1−i2 through CD as shown below:
Now the question is simplified. We can use Kirchhoff’s law to calculate i1 and i2 in terms of V and apply V=iReq to get Req.
Let’s do that. Refer the following diagram:
Loop ACDFA
−10i1−5(i1−i2)+5i2=0−15i1+10 i2=0i1=23i2
Loop ACEBGHA:
−10i1−5 i2+V=010 i1+5i2=V
Solving both the equations in finding i1 and i2 in terms of V, we get
i2=335 V
i1=235 V
i=i1+i2=535 V=17 V
Which implies, Vi=7Ω
Thus equivalent resistance between A and B is 7Ω.
