Question

a 120v ac circuit contains 10 ohm redistance and 30 ohm resistance in series what would be the average power in circuit

Answer 1

We know that I =V/U.

v=120
R s=Rs1+Rs2(series connection)
IE 30+10=40

substitute
I=120/40=3

that is 3A

Answer 2

Given,

Total voltage supply across a circuit = 120 V

The circuit contains a 10 ohms resistor and a 30 ohms resistor in a series arrangement.

To find,

The average power in the circuit.

Solution,

We can simply solve this numerical problem by using the following process:

Mathematically,

The average power in a circuit

= (voltage supply across a circuit)×(current flowing through the circuit)

{Statement-1}

If a set of resistors are arranged in a series type of arrangement, then the equivalent resistance of the arrangement is equal to the total of the individual resistances. The total current across all the resistors, arranged in series type, is equal.

{Statement-2}

As per ohms law,

For a voltage V across the conductor, current I flowing through the conductor and resistance R provided by the conductor to the flow of current,

V = RI

{Statement-3}

Now, according to the question and statement-2;

The equivalent resistance of the circuit

= sum of the individual resistances arranged in series type

= 10 ohms + 30 ohms

= 40 ohms

Now, according to the question and statement-3;

The total current flowing through the circuit

= (voltage supply across the circuit)/(equivalent resistance of the circuit)

= (120 V)/(40 ohms)

= 3 A

Now, according to question and statement-1;

The average power in the circuit

= (voltage supply across a circuit)×(current flowing through the circuit)

= (120 V)×(3 A)

= 360 W

Hence, the average power in the circuit is equal to 360 Watts.