In the circuit shown- the reading of the Ammeter is doubled after the switch is closed- Each resistor has a resistance- 1 - and the ideal cell has an e-m-f- -10V- Then the ammeter has a resistance equal to

The correct option is D 2 Ω Let resistance of Ammeter be r . Case 1 I_1=V2+r Case 2 I_2=Vr Given, I_2=2I_1 10r=2× 102+r ...

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Byju's Answer

Standard XII

Physics

Ohm's Law in Scalar Form

In the circui...

Question

In the circuit shown, the reading of the Ammeter is doubled after the switch is closed. Each resistor has a resistance= $1 Ω$ and the ideal cell has an $e . m . f . = 10 V$ . Then the ammeter has a resistance equal to

2 Ω

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1 Ω

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2.5 Ω

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5 Ω

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Solution

The correct option is D $2 Ω$
Let resistance of Ammeter be $r$ .
$Case 1$

$I_{1} = \frac{V}{2 + r}$

$Case 2$

$I_{2} = \frac{V}{r}$

Given,
$I_{2} = 2 I_{1}$

$\frac{10}{r} = \frac{2 \times 10}{2 + r}$

$2 r = 2 + r$

$r = 2 Ω$

Option $A$ is the correct answer

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In the circuit shown, the reading of the Ammeter is doubled after the switch is closed. Each resistor has a resistance=1 Ω and the ideal cell has an e.m.f.=10V. Then the ammeter has a resistance equal to

The correct option is D 2 ΩLet resistance of Ammeter be r.Case 1I1=V2+rCase 2I2=VrGiven,I2=2I110r=2×102+r2r=2+rr=2ΩOption A is the correct answer

In the circuit shown, the reading of the Ammeter is doubled after the switch is closed. Each resistor has a resistance= $1 Ω$ and the ideal cell has an $e . m . f . = 10 V$ . Then the ammeter has a resistance equal to

The correct option is D $2 Ω$
Let resistance of Ammeter be $r$ .
$Case 1$

$I_{1} = \frac{V}{2 + r}$

$Case 2$

$I_{2} = \frac{V}{r}$

Given,
$I_{2} = 2 I_{1}$

$\frac{10}{r} = \frac{2 \times 10}{2 + r}$

$2 r = 2 + r$

$r = 2 Ω$

Option $A$ is the correct answer