In a circuit, there are two resistors of 3 Ω and an unknown resistor and they are connected in series across a 12 V battery. If the potential difference across the unknown resistor is 6 V then the equivalent resistance of the circuit is [[1]] Ohms.

When two resistors of

3 Ω

and an unknown resistor are connected in series across a

12 V

battery and if the potential difference across the unknown resistor is

6 V

then the equivalent resistance of the circuit is

12 O h ms

.
Two resistors of resistances

3 Ω

and an unknown resistance connected in series are given.
The potential difference across the unknown resistance is given as

6 V

.
Let the potential difference across the

3 Ω

resistance be

V_{1}

. Therefore, the total resistance is

12 = 6 + V_{1} + V_{1}

\Rightarrow 6 = {2 V}_{1}

{\Rightarrow V}_{1} = 3 V

Hence, the resistance across the

3 Ω

resistance is

3 V

.
Now, from the Ohm’s law, the current in the resistor is given as

I = \frac{V}{R}

On substituting the given values, we get the current in the resistor

3 Ω

I = \frac{3}{3}

\Rightarrow I = 1 A

Therefore, the current in the

3 Ω

resistor is

1 A

, as all the resistors are connected in series connection and hence, the current in the unknown resistor is

1 A

.
The potential difference across the unknown resistor is

6 V

and the current flowing through it is

1 A

. Therefore, the value of resistance is

R = \frac{V}{I}

\Rightarrow R = \frac{6}{1}

\Rightarrow R = 6 Ω

Since, the resistors are connected in series therefore the equivalent resistance is,

R_{S} = 3 + 3 + 6

\Rightarrow R_{S} = 12 Ω

Hence, the equivalent resistance of the circuit is

12 Ω

In a circuit, there are two resistors of $3 Ω$ and an unknown resistor and they are connected in series across a $12 V$ battery. If the potential difference across the unknown resistor is $6 V$ then the equivalent resistance of the circuit is [[1]] $O h ms$ .