Question

A battery of 9 V is connected in series with resistors of 2ohm,3ohm,4ohm and 5ohm respectively.
Calculate the current flowing in the circuit.
Draw the circuit diagram.​

Answer 1

Answer:

0.64 A

Explanation:

Since resistors are connected in series,

$\\\tt Resistance_{net} = (2 + 3+4+5)\Omega\\\\\tt \implies 14 \Omega$

Using Ohm's law

V=IR

$\\\tt I = \dfrac{V}{R} \\\\\tt = \dfrac{9}{14} A\\\\\tt = 0.64 A$

Consider the attachment for the diagram:

Answer 2

Given :-

A battery of 9 V is connected in series with resistors of 2ohm,3ohm,4ohm and 5ohm respectively.

To Find :-

Current

Solution :-

They are in series so,

$\sf R_n = R1+R2+R3+R4$

$\sf R = 2+3+4+5$

$\sf R = 14\Omega$

We know that

$\bf {V=IR}$

$\sf 9 = I \times14$

$\sf\dfrac{9}{14} = I$

$\sf 0.64 = I$