In the given circuit,it is observed that the current $I$ is independent of the value of the resistance $R_6$. Then the resistance values must satisfy

  • A
    $R_1 R_2 R_5 = R_3 R_4 R_6$
  • B
    $\frac{1}{R_5} + \frac{1}{R_6} = \frac{1}{R_1 + R_2} + \frac{1}{R_3 + R_4}$
  • C
    $R_1 R_4 = R_2 R_3$
  • D
    $R_1 R_3 = R_2 R_4 = R_5 R_6$

Explore More

Similar Questions

The readings of ammeters $A_1, A_2$ and $A_3$ will be respectively:

In the circuit shown below,the ammeter reading is zero. Then the value of the resistance $R$ is (in $Omega$)

$A$ Wheatstone bridge is used to determine the value of an unknown resistance $X$ by adjusting the variable resistance $Y$ as shown in the figure. For the most precise measurement of $X$,the resistances $P$ and $Q$:

Four ammeters with identical internal resistances $r$ and a resistor of resistance $R$ are connected to a current source as shown in the figure. It is known that the reading of the ammeter $A_1$ is $I_1 = 3\ A$ and the reading of the ammeter $A_2$ is $I_2 = 5\ A$. Determine the ratio of the resistances $R/r$.

Difficult
View Solution

In the following electrical network, the value of $I$ is (in $\text{ A}$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo