In an experiment according to set up, when E1=12 V and internal resistance zero, e=2 V . The galvanometer reads zero, then X would be-

Question

In an experiment according to set up, when E1=12 V and internal resistance zero, e=2 V . The galvanometer reads zero, then X would be- <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-yfkukurvcedduudw.jpg" /> (A) 200 (B) 500 (C) 100 (D) 10

Tardigrade Admin · Accepted Answer

Voltage across X is 2 v text So ((E1/500+X)) X=2 ⇒((12/500+X)) X=2 ⇒ 12 X=1000+2 X ⇒ 10 X=1000 X =100 Ω

Q. In an experiment according to set up, when $E_{1} = 12 V$ and internal resistance zero, $e = 2 V$ . The galvanometer reads zero, then $X$ would be-

$200$

$500$

$100$

$10$

Voltage across $X$ is $2 v$
$So (\frac{E _{1}}{500 + X}) X = 2$
$\Rightarrow (\frac{12}{500 + X}) X = 2$
$\Rightarrow 12 X = 1000 + 2 X$
$\Rightarrow 10 X = 1000$
$X = 100 Ω$

Q. In an experiment according to set up, when E1​=12V and internal resistance zero, e=2V . The galvanometer reads zero, then X would be-

200

500

100

10