The effective resistance between points A and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given circuit is a balanced Wheatstone bridge because the ratio of resistances in the arms is equal $(10/10 = 10/10)$.In a balanced Wheatstone

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The given circuit is a balanced Wheatstone bridge because the ratio of resistances in the arms is equal $(10/10 = 10/10)$.In a balanced Wheatstone bridge,the potential difference across the central diagonal resistor is zero,so no current flows through it.We can remove the central $10\,\Omega$ resistor from the circuit.Now,the upper two $10\,\Omega$ resistors are in series,giving an equivalent resistance of $10 + 10 = 20\,\Omega$.The lower two $10\,\Omega$ resistors are also in series,giving an equivalent resistance of $10 + 10 = 20\,\Omega$.These two branches are now in parallel between points $A$ and $B$.The total effective resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$.Therefore,$R_{eq} = 10\,\Omega$.

The effective resistance between points $A$ and $B$ is ............... $\Omega$.

Similar Questions

Calculate the potential difference between $A$ and $B$ in the given circuit in $volt$.

For the network shown in the figure,the value of the current $i$ is:

In the given circuit,the current $I_1$ is .............. $A$.

The four arms of a Wheatstone bridge have resistances as shown in the figure. $A$ galvanometer of $15\, \Omega$ resistance is connected across $BD$. Calculate the current through the galvanometer when a potential difference of $10\, V$ is maintained across $AC$.

In the given circuit, the values of $I_1, I_2, I_3$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module