Question

If 2 resistors of 5kΩ and 7kΩ are connected in series across a 15V battery, then the current flowing through each of the resistance should be


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Answer 1

Explanation:

Aa they are in series

R= 5+7= 12kohm

V= IR

I = V/R

= 15/12

= 1.25mA

Hence D is the ans

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Answer 2

Answer:

The current flowing through each of the resistance should be 1.25mA.

Explanation:

As the resistors are connected in the series same amount of current will flow through them.

The resistance of the first resistor(R₁)=5kΩ

The resistance of the second resistor(R₂)=7kΩ

The potential of the battery(V)=15 volt

The equivalent resistance of the resistors is,

$R=5k\Omega+7k\Omega$

$R=12k\Omega$   (1)

From the Ohms law, we have,

$V=IR$     (2)

By inserting the values in equation (2) we get;

$15=I\times 12\times 10^3$

$I=\frac{15}{12\times 10^3}$

$I=1.25\times 10^-^3A$

$I=1.25mA$

Hence, the current flowing through each of the resistance should be 1.25mA.

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