English
Karnataka Board PUCPUC Science Class 11

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

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

(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

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

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

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 10: Electric Current in Conductors - Exercises [Page 200]

APPEARS IN

HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 10 Electric Current in Conductors
Exercises | Q 41 | Page 200
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×