Question

3 resistors 5 8 12 ohm each connected in series circuit battery has 3 cells 2 volts each find the current​

Answer 1

We have:-

• Three resistors 5 Ω, 8 Ω and 12 Ω
• Three cells of 2 V each.

To FinD:-

• Current flowing through the conductor$?$

Explanation:-

The resistors are connected in series, which means the equivalent resistance will be the sum total of all resistances.

• Req. = R1 + R2 + R3

➝ Req. = 5 Ω + 8 Ω + 12 Ω

➝ Req. = 25 Ω

Now finding the total voltage / potential difference provided by the source i.e. battery:

• Veq. = n × V

It is because all the potential differences provided by the cells are equal. n is the no. of cells in the battery.

➝ Veq. = 3 × 2 V

➝ Veq. = 6 V

Now finding the current by Ohm's law:

➝ Veq. = IReq.

➝ 6 V = I × 25 Ω

➝ I = 6 / 25 A

➝ I = 0.24 A

Hence:-

• The current flowing through the wire or the circuit is 0.24 A.

Answer 2

Explanation:

Given :

• resistors 5 8 12 ohm each connected in series circuit battery has 3 cells 2 volts

To Find :

• find the currents

Solution :

Potential difference = 3 × 2 = 6 V

Net resistance = 5 + 8 + 12 = 25 ohm

Finding the Current (Ammeter reading) –

V = I × R

6 V = I × 25

I = 6/25

l = 0.24 A

• Hence Ammeter reading = current = 0.24 A

More to know :

• An Ammeter is always connected in series.

• A Voltmeter is always connected in parallel to the resistor.

• An Ammeter is used to find out the current.

• A Voltmeter is used to find out the potential difference across the resistor.