Question

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State the reason properly...

Why is it 0.6A instead of 1.2A??


Please help! ​

Answer 1

Answer:

$\bigstar\;{\underline{\underline{\textbf{Situation - 1}}}$

• When k₁ is closed :

The current is only passing through 5Ω resistor.

Let the :

The Lamp of resistance be - R

There is zero current passing through 10Ω.

According to the diagram,

5Ω and RΩ are in series.

Therefore,

Req = 5Ω + RΩ.

We know that :

Ohm's Law :

$\bigstar\;\boxed{\sf V = IR}$

So,

As given,

i = 0.6A

⇒ 6 = 0.6 × ( 5 + R)

⇒ 10 = 5 + R

⇒ R = 5Ω

∴ The resistance of lamp is 5Ω.

$\rule{300}{1.5}$

$\bigstar\;{\underline{\underline{\textbf{Situation - 2}}}$

• When k₂ is closed :

Let the :

i current flows through the circuit.

So,

Req = 10Ω × 10Ω/(10Ω + 10Ω) = 5Ω

∵ The two resistors each of resistance 5Ω are joined in series and and both are joined with 10Ω resistor in parallel.

So,

⇒ Req = 5Ω

The current through the circuit is i :

⇒ V/R

⇒ 6/5

⇒ 1.4A

Because,

10Ω resistor and (5Ω + 5Ω) resistors are connected in parallel.

We also know,

Potential in parallel combination is same.

Therefore,

Current will be half due to each of resistors have same resistance. i.e 10Ω.

Now,

Current through 5Ω resistor is 0.7A

Current through lamp is 0.7A

Potential difference across lamp :

⇒ 0.7 × 5

⇒ 3.5 Volt

Answer 2

Answer:

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