The reading of the ideal ammeter will be (Resistance of ideal ammeters is zero)

Question

The reading of the ideal ammeter will be (Resistance of ideal ammeters is zero) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-qtftmuxll8ed.png" /> (A) 5 / 6 ampere (B) 6 / 5 ampere (C) 3 / 2 ampere (D) 2 / 3 ampere

Tardigrade Admin · Accepted Answer

Re q=3 ||6 + 2 = 2 + 2 = 4 Ω Current from Battery, I=(V/Re q)=(10/4)=2.5A <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-qtftmuxlo5gk.png" /> I1=2.5× (3/3 + 6)=(5/6) A

Q. The reading of the ideal ammeter will be (Resistance of ideal ammeters is zero)

$5/6 am p ere$

$6/5 am p ere$

$3/2 am p ere$

$2/3 am p ere$

$R_{e q} = 3∣∣6 + 2 = 2 + 2 = 4Ω$
Current from Battery, $I = \frac{V}{R _{e q}} = \frac{10}{4} = 2.5 A$

$I_{1} = 2.5 \times \frac{3}{3 + 6} = \frac{5}{6} A$

Q. The reading of the ideal ammeter will be (Resistance of ideal ammeters is zero)

5/6ampere

6/5ampere

3/2ampere

2/3ampere

Req​=3∣∣6+2=2+2=4Ω Current from Battery, I=Req​V​=410​=2.5A I1​=2.5×3+63​=65​A

$5/6 am p ere$

$6/5 am p ere$

$3/2 am p ere$

$2/3 am p ere$

$R_{e q} = 3∣∣6 + 2 = 2 + 2 = 4Ω$
Current from Battery, $I = \frac{V}{R _{e q}} = \frac{10}{4} = 2.5 A$

$I_{1} = 2.5 \times \frac{3}{3 + 6} = \frac{5}{6} A$