Question

If two resistors of 3 kW and 9 kW are connected in
series across a 15 V battery, then current flowing
through each of resistance should be..
plz be fast​

Answer 1

• 0.00125 Ampere

Explanation:

Given:-

• Resistance of Resistor ,R1 = 3 KΩ

• Resistance of Resistor ,R2 = 9 KΩ

• Potential Difference ,V = 15 v

To Find:-

• Current flowing through Resistor ,I

Solution:-

Firstly we calculate the equivalent resistance of resistors.As it is given that the Resistors are connected in series .

• Rs = R1 + R2 + ..

Where,

• Rs Equivalent resistance
• R1 Resistance of Resistor
• R2 Resistance of Resistor

Substitute the value we get

→ Rs = 3 KΩ + 9 KΩ

→ Rs = 12 KΩ

→ Rs = 12×1000 = 12000 KΩ

• Equivalent Resistance in series is 12KΩ.

Now, We have to calculate the Current flowing through the Resistors . Using Ohm's Law

• V = IR

Where,

• V denotes Voltage
• I denotes Current
• R denote Resistance

Substitute the value we get

→ 15 = I × 12000

→ I = 15/12000

→ I = 0.00125 A

• Hence, the current flowing through the Resistors is 0.00125 Ampere.

Answer 2

$\Large\bf{\color{indigo}GiVeN,}$

• Two resistors of $\bf {3k\Omega}$ and $\bf {9k\Omega}$ are connected in series.

• 15 V battery is connected across the combination of these two resistors.

$\Large\bf\pink{We\:know\:that,}$

☆ When two resistors are connected in series, then the equivalent resistance is

$\red\bigstar\:\:\bf\purple{R_{eq}\:=\:R_1\:+\:R_2\:}$

$\bf\blue{Where,}$

• $\bf{R_1\:=\:3k\Omega}$

• $\bf{R_2\:=\:9k\Omega}$

$:\implies\:\:\bf{R_{eq}\:=\:3k\Omega\:+\:9k\Omega\:}$

$:\implies\:\:\bf\green{R_{eq}\:=\:12000\Omega\:}$

$\Large\bf\orange{Again\:we\:have,}$

$\green\bigstar\:\:\bf\blue{Current\:=\:\dfrac {Voltage}{R_{eq}}\:}$

$\longmapsto\:\:\bf{Current\:=\:\dfrac {15}{12000}\:}$

$\longmapsto\:\:\bf{Current\:=\:\dfrac {1}{800}\:}$

$\longmapsto\:\:\bf\purple{Current\:=\:0.00125\:A\:}$

$\Large\star$ As we know that in series combination "same current passes through each resistance".

$\Large\bold\therefore$ Current flowing through each of resistance should be "0.00125 A".