Question

Three resistors each of 100 are connected in series to a battery of potential difference 150 V. The current flowing through it is _____A​

Answer 1

Three resistors each of resistance 100Ω are connected in series to a battery of potential difference 150 V.

R1 = R2 = R3 = 100Ω

--------| R1 |-------| R2 |-------| R3 |----------

For series combination:

Rs = R1 + R2 + R3

For parallel combination:

1/Rp = 1/R1 + 1/R2 + 1/R3

As said in question, three resistors are connected in series. So,

Rs = R1 + R2 + R3

Substitute value of R1, R2 and R3 in the above formula,

Rs = 100 + 100 + 100

Rs = 100(1 + 1 + 1)

Rs = 100(3)

Rs = 300

Req = Rs

Req = 300Ω

Ohm's law: The potential difference i.e. V across the ends of the wire is directly proportional to the current i.e. I flowing through that wire provides the temperature remains constant.

V ∝ I

V = IR (here, R is constant)

From ohm's law we can say that,

V = IR (where R is resistance)

Substitute the known values in the above formula,

150 = I × 300

150/300 = I

0.5 = I

Therefore, the current flowing through it is 0.5 A.

Answer 2

Answer:

0.5 Amperes

Explanation:

Given :

• Three resistors of each 100 ohms are connected in series to each other

• Potential difference across the resistor = v = 150 Volts

To find :

• The current flowing through the circuit

Using the ohms law:

V=I×R

Equivalent resistance = 100+100+100

Equivalent resistance = 300 ohms

150=I×300

I=150/300

I=1/2

I=0.5 Amperes

The current passing through the circuit is equal to 0.5 Amperes