Question

Q) In the figure given below , there three resistors with resistances 2ohm, 3ohm and 5ohm respectively are connected in series with 10V battery. Calculate the equivalent resistance and current that passes through each resistor in the given network ?
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Answer 1

Answer:

(3p+q)(p2q−2pq)=3p3q−6p2q+p2q2−2pq2

Answer 2

Answer:

The equivalent resistance of the circuit is 10 Ω and current of 1 A is following through each resistor.

Explanation:

The equivalent resistance of the resistor in series is given as,

$R_{eq}=R_1+R_2+R_3$

Substituting all known all known values,

[tex]R_{eq}=2 \Omega+3\Omega+5\Omega\\ =10 \Omega[/tex]

According to Ohm's law,

$V=IR_{eq}$

Therefore, the current flowing in the circuit is given as,

$I=\frac{V}{R_{eq}}$

Substituting all known values,

[tex]I=\frac{10\text{ V}}{10\Omega}\\ =1\text{ A}[/tex]

As the resistors are connected in series connection same current (1 A) will flow through them..

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