Question

the value of 2 resistor in series is in the ratio of 1 ratio 4 and a 10 ampere current flow in the circuit if these two resistors are combined in a parallel circuit then calculate the ratio of the current flowing through the resistors​

Answer 1

Answer:

Explanation:

Resistors in Series

Resistors are said to be connected in series when they are daisy chained together in a single line resulting in a common current flowing through them

Answer 2

Answer:

Correct question :

The ratio of resistance of two resistors

A and B joined in series combination is 1 : 4.The current flowing through the circuit is 10A.If these resistances be joined in parallel combination then find the ratio of currents that would flow through them

Explanation:

$</p><p>Let \: R_{1} and \: R _{2} \: be \: resistance \: of \: resistors \: A \: and \: B \: respectively.$

$Given \: \: \: \: \: \: \: \:R _{1} : R_{2} = 1 : 2 \: \: \: \: \: \: ....(1)$

In parallel combination potential difference across each resistor is same but current flowing through them are different.

$i.e \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: V=IR$

But potential is constant

$∴ \: \: \: \: \: \: \: \: \: \: IR=Constant .$

$⇒ \: \: \: \: \: \: \: \: I ∝ \frac{1}{R}$

$∴ \: \: \: \: \: \: \: \: \: \: \frac{ I_{1} }{ I_{2} } = \frac{R _{2} }{ R_{1} } = \frac{4}{1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [using \: eqn.(i)]$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: I _{1} = 4I_{2}$

$Given \: \: \: \: \: \: \: \:I _{1} + I _{ 2 } = 10 A$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: 4I_{2} + I_{2} = 10$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: I_{2} = 2A$

$and \: \: \: \: \: \: \: \: \: \: \: \: I _{1} = 4 I _{2} = 8 A$

$Therefore, \: \: \: \: \: \: \: I _{1} : I_{2} = 8A : 2A$

$or \: \: \: \: \: \: \: \: \: \: \: \: \: I _{1} : I_{2} = 4 : 1.$