Question

A battery of 12V is connected in series with resister of 0.2ohm , 0.3ohm , 0.4ohm , 0.5ohm , and 12ohm. How much current will flow through the 0.3ohm resister

Answer 1

Given that, A battery of 12V is connected in series with resister of 0.2ohm, 0.3ohm, 0.4ohm, 0.5ohm, and 12ohm.

R1 = 0.2Ω, R2 = 0.3Ω, R3 = 0.4Ω, R4 = 0.5 and R5 = 12Ω

Rs = R1 + R2 + R3 + R4 + R5

Rs = (0.2 + 0.3 + 0.4 + 0.5 + 12)Ω

Rs = (1.4 + 12)Ω

Rs = 13.4Ω

From Ohm's Law, we can say that,

V = IR

Given V is 12 V and from the above calculations R = 13.4Ω.

12 = I × 13.4

12/13.4 = I

0.89 = I

We have to find how much current will flow through the 0.3Ω resistor.

{ All resistors are connected in series. So, the current will be same in all resistors. }

I = I1 = I2 = I3 = I4 = I5

Therefore, 0.89A of current will flow through the 0.3Ω resistor.

Answer 2

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Given :-

A battery of 12 V is connected in series with resister of 0.2 Ω , 0.3 Ω , 0.4 Ω, 0.5 Ω , and 12 Ω.

To find :-

The amount of current that will flow through the 0.3 Ω resistor.

Solution :-

$\sf{\displaystyle{R_1=0.2 \Omega},\ \displaystyle{R_2=0.3 \Omega},\ \displaystyle{R_3=0.4\Omega},\ \displaystyle{R_4=0.5\Omega},\ and\ \displaystyle{R_5=12\Omega}.}$

As the resistors are connected in series, so,

$\sf{\displaystyle{R_S=R_1+R_2+R_3+R_4+R_5}}$

$\displaystyle{\sf{\implies R_S=(0.2+0.3+0.4+0.5+12)\Omega}}\\\\\displaystyle{\sf{\implies R_S=13.4 \Omega}}$

By Ohm's Law*, we know :

$\bold{\huge\sf{V=IR}}$

Potential difference = 12 V        ...(given)

Putting the known values in the equation :

12 = I*(13.4)

⇒ I = 12/13.4

⇒ I = 0.89 A

As all the resistors are connected in series, so the current flow will be the same throughout all the resistor. i.e.,

$\displaystyle{\sf{I_1=I_2=I_3=I_4=I_5=I}}$

So, the current that flows through all the resistors is equal. So, the current that flows through the 0.3 Ω resistor is 0.89 A.

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Ohm's Law* :- Ohm's law states that the voltage(V) across a resistor is directly proportional to the current(I) flowing through the resistance(R).