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Find the Equivalent Resistance of the Circuits Shown in the Figure Between the Points A And B. Each Resistor Has Resistance R. - Physics

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प्रश्न

Find the equivalent resistance of the circuits shown in the figure between the points a and b. Each resistor has resistance r.

योग

उत्तर

(a) The simplified circuit can be drawn as shown below.

Here cdeo forms a balanced Wheatstone bridge; therefore, branch od will become become ineffective.

The simplified circuit will then be as shown below.

The equivalent resistance between points c and e,

\[R_{cd} = \frac{2r \times 2r}{4r} = r\]

The equivalent resistance between a and b,

\[R_{ab} = \frac{r \times r}{2r} = \frac{r}{2}\]

 

(b)

Let Reff  be the effective resistance of the circuit.Then, from the symmetry of the circuit, we can assume that the current moving along CO enters OB and the current moving along EO enters OD.

Current on CO = current on OB

Current on EO = current on OD

So, the circuit can be simplified as shown below.

From the simplified circuit diagram, effective resistance of the upper half of the circuit will  be

\[\left[ \left[ r + \left( \frac{2 r^2}{3r} \right) + r \right] = 2r + \left( \frac{2r}{3} \right) = \frac{8}{3}r \right]\]

\[ \Rightarrow R_{eff} = \frac{\frac{8r}{6} \times 2r}{\frac{8r}{6} + 2r}\]

\[ = 8 r^2 \times \frac{2}{20r} = \frac{8r}{10} = \frac{4r}{5}\]

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अध्याय 10: Electric Current in Conductors - Exercises [पृष्ठ २००]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 10 Electric Current in Conductors
Exercises | Q 41 | पृष्ठ २००

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  • Reason (R): All the resistors are connected in parallel.

 


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