Reading of ammeter is 2A . Find value of R (in W ) .

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Reading of ammeter is 2A . Find value of R (in W ) . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-qqxh7pcrsq3ddolh.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-qqxh7pcrsq3dv64v.jpg" /> VP=2× 2=4V ⇒ Current through 3Ω=(10 - 4/3)=2A Means no current through 5Ω . ⇒ Wheatstone bridge. ⇒ (R'/3)=(4/2)⇒ R'=6Ω

Q. Reading of ammeter is $2 A$ . Find value of $R$ (in $W$ ) .

$V_{P} = 2 \times 2 = 4 V$
$\Rightarrow$ Current through $3Ω = \frac{10 - 4}{3} = 2 A$
Means no current through $5Ω$ . $\Rightarrow$ Wheatstone bridge. $\Rightarrow \frac{R ^{^{'}}}{3} = \frac{4}{2} \Rightarrow R^{^{'}} = 6Ω$

Q. Reading of ammeter is 2A . Find value of R (in W ) .

VP​=2×2=4V ⇒ Current through 3Ω=310−4​=2A Means no current through 5Ω . ⇒ Wheatstone bridge. ⇒3R′​=24​⇒R′=6Ω

Q. Reading of ammeter is $2 A$ . Find value of $R$ (in $W$ ) .

$V_{P} = 2 \times 2 = 4 V$
$\Rightarrow$ Current through $3Ω = \frac{10 - 4}{3} = 2 A$
Means no current through $5Ω$ . $\Rightarrow$ Wheatstone bridge. $\Rightarrow \frac{R ^{^{'}}}{3} = \frac{4}{2} \Rightarrow R^{^{'}} = 6Ω$