Question

Three `resistances, each equal to 5ohm,are connected in series with a cell of 1.2v .find the magnitude of current in the circuit

Answer 1

Three resistors, each of resistance 5 ohm are connected in series with a cell emf of 1.2 V.

From above data we have, R1 = R2 = R3 = 5 ohm and V = 1.2V

We have to find the magnitude of current in the circuit.

For series combination:

Rs = R1 + R2 + R3

For parallel combination:

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

As per given condition, they are connected in series. So,

Rs = R1 + R2 + R3

Rs = 5( 1+ 1+ 1)

Rs = 5(3)

Rs = 15 ohm

Therefore, the value of resistance is 15 ohm.

From ohm's law we can say that,

V = IR

Substitute value of R = 15 ohm and V = 1.2V

1.2 = I × 15

1.2/15 = I

0.08 = I

Therefore, the magnitude of current in the circuit is 0.08 A.

Answer 2

Given :

• Three resistors each of 5Ω are connected in parallel.
• Potential Difference (v) = 1.2 v

To Find :

We have to find the amount of current flowing through it.

Explanation :

We know the formula to calculate the resistance in series.

⟹ Rs = R1 + R2 + R3

→ Rs = 5 + 5 + 5

→ Rs = 15

∴ Total resistance = 15 Ω

$\rule{200}{2}$

Now,

⟹ V = IR

→ 1.2 = I * 15

→ 1.2/15 = I

→ I = 0.08

∴ Current flowing through it is of 0.08 A.