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The current flowing through the circuit shown in the figure is [[1]] $A$ and the potential difference across the $10 Ω$ resistor is [[2]] $V$ .

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Detailed Solution

The current flowing through the circuit shown in the figure is

0.1 A

and the potential difference across the

10 Ω

resistor is

1 V

.
Two resistors of

10 Ω

and

20 Ω

connected in series and a voltage supply of

3 V

is given.
Since, the resistors

10 Ω

and

20 Ω

are in series, the equivalent resistance is given by the sum of their resistances, i.e.

R_{S} = R_{1} + R_{2}

{\Rightarrow R}_{S} = 10 + 20

\Rightarrow R_{S} = 30 Ω

From Ohm’s law, the current flowing in the circuit is given as

I = \frac{V}{R_{S}}

\Rightarrow I = \frac{3}{30}

\Rightarrow I = 0.1 A

The potential difference across the

10 Ω

resistor is calculated as

V = IR

\Rightarrow V = \frac{1}{10} \times 10

\Rightarrow V = 1 V

Hence, the current flowing through the circuit is

0.1 A

and the potential difference across the

10 Ω

resistor is

1 V

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