Question

two resistors each of 3 ohm and unknown resistors are connected in series across a 12 volt battery is potential across 3 ohm resistor the current through unknown resistor (c)equivalent resistance of the circuit.​

Answer 1

Answer:

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Explanation:

Given Data :

There are 2 resistors, one is of 3 ohm, i.e \sf{R_{1}=3\:Ohms}R1=3Ohms and \sf{R_{2}}R2 is unknown.

These resistors are connected in series.

Voltage is given as 12v, and Voltage drop across \sf{V_{2}=6v}V2=6v

To Find :

Potential Difference across 3 ohm resistance.

The current through the unknown resistor

Equivalent Resistance of the circuit.

As it is given that it is series connection,

\boxed{ \sf{v = v_{1} + v_{2}}}v=v1+v2

Plugging the values,

\sf{12 = v_{1} + 6}12=v1+6

\sf{ \green{v_{1} = 6v }}v1=6v

The voltage drop across \sf{V_{1}}V1 is 6v.

Potential drop across 3 ohm resistor :

Using Ohm's Law,

\boxed{ \sf{V = IR}}V=IR

Where,

\sf{v_{1} = 6}v1=6 , I should be found, and \sf{R_{1}=3}R1=3

\sf{6 = I \times 3}6=I×3

\sf{ \blue{I = 2A}}I=2A

We found the current for first resistor, i.e \sf{i_{1}}i1 which is 2 Ampere.

As we know that in a series connection the current is same for both the resistors,

So \sf{I_{2}=2A}I2=2A

The current flowing through the unknown resistor is 2A.

Let us now find the value of unknown resistor in order to find the equivalent resistance of the circuit.

Using Ohm's Law,

\boxed{ \sf{V = IR}}V=IR

The voltage for second resistor is 6v, Current flowing is 2A, Using this let's find the value of \sf{R_{2}}R2

Plugging in the values we get,

\sf{6 = 2 \times R_{2}}6=2×R2

\sf{ \red{R_{2} = 3 \: ohms}}R2=3ohms

So the resistance of the unknown resistor is 3 ohm.

Let us now find the equivalent resistance.

In a series connection,

\boxed{ \sf{R = R_{1} + R_{2}}}R=R1+R2

\sf{R = 3 + 3}R=3+3

\sf{R = 6 \: ohms}R=6ohms

The equivalent resistance of the given circuit is 6 ohms.

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Note : Don't forget to write the SI units (warna ½ mark kat jayega T_T xD)

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