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

Question

The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15 Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC . <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/596909a74b39f98a777d333607abe3b7-.png" /> (A) 2.44 μ A (B) 2.44 mA (C) 4.87 mA (D) 4.87 μ A

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/6933a41001e51c835f8d8d199a83131f-.png" /> (x-10/100)+(x-y/15)+(x-0/10)=0 53 x-20 y=30 ldots . .(1) (y-10/60)+(y-x/15)+(y-0/5)=0 17 y-4 =10 ldots ldots(2) on solving (1) &(2) x=0.865 y=0.792 Δ V=0.073 R=15 Ω i=4.87 mA

Q. The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of $15 Ω$ resistance is connected across $B D$ . Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across $A C .$

$2.44 μ A$

$2.44 m A$

$4.87 m A$

$4.87 μ A$

$\frac{x - 10}{100} + \frac{x - y}{15} + \frac{x - 0}{10} = 0$
$53 x - 20 y = 30 \dots .. (1)$
$\frac{y - 10}{60} + \frac{y - x}{15} + \frac{y - 0}{5} = 0$
$17 y - 4 = 10 \dots\dots (2)$
on solving $(1) & (2)$
$x = 0.865$
$y = 0.792$
$Δ V = 0.073 R = 15Ω$
$i = 4.87 m A$

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

2.44μA

2.44mA

4.87mA

4.87μA