Question

In the adjacent figure bridge is balanced, the current flowing through 2$\rho$ resistance is

• A. $\frac{r_{1}}{r_{2}}\frac{\rho}{2}$
• B. $\frac{r_{2} - r_{1}}{r_{1}r_{2}}\frac{\rho}{4\pi}$
• C. $\frac{r_{1}r_{2}}{r_{2} - r_{1}}\frac{\rho}{4\pi}$
• D. $2\Omega,4\Omega$

Answer 1

Answer: (A)

$\frac{r_{1}}{r_{2}}\frac{\rho}{2}$

Answer 2

: Let I be the current flowing through PQR then the current flowing through PSR is (2 –I) A.

Now since, galvanometer shows no deflection then the voltage through arm PQR is same as the voltage through arm PSR

$I(4 + 2) = (2 - I)(10 + 5)$

$6I\backslash 30 - 15I$

$21I = 30 \Rightarrow I = \frac{30}{21} = \frac{10}{7}A$

Hence the current through $2\Omega$resistor is $\frac{10}{7}A$