Question

Three identical cells,each of emf 2V and internal resistance 0.2 ohms are connected in series to an external resistor of 7.4 ohms. Calculate the current in the circuit.

Answer 1

Easy here you go

I hope this helps

Answer 2

AnswEr :

From the Question,

• Emf (E) = 2 V

• Internal Resistance (r) = 0.2 Ω

• Resistance (R) = 7.4 Ω

Three cells of 2V and 0.2 each are connected in series.

Consider two cells connected to eachother.

The total potential difference would be :

$\sf \: V = V_1 + V_2$

Total Potential Difference in a cell is given as :

$\sf \: V = E - IR$

Thus,

$\sf \implies E - Ir = E_1 - Ir_1 + E_2 + Ir_2 \\ \\ \implies \sf E_{eq} = E_1 + E_2 \ and \ r_{eq} = r_1 + r_2$

Therefore,

For three cells each of emf 2V,

$\sf \: \: E = 2 + 2 + 2 = 6 \: V$

For three cells with internal resistance 0.2 Ω,

$\sf \: r = 0.2 + .0.2 + 0.2 = 0.6 \: \Omega$

We know that,

$\star \ \boxed{\boxed{\sf I = \dfrac{E}{r + R} }}$

Implies,

$\longrightarrow \sf \: l = \dfrac{6}{06 + 7.4} \\ \\ \longrightarrow \sf \: l = \dfrac{6}{8} \\ \\ \longrightarrow \large{ \underline{\boxed{ \: \sf \: l = 0.75 \: A}}}$