A battery of e.m.f. 15 V and internal resistance 2 Ω is connected to two resistors of resistances 4 Ω and 6 Ω joined in series. Find the electrical energy spent per minute in 6 Ω resistor.

(i) Given,

e.m.f. (V) = 15 V

Internal resistance = 2 Ω

Resistors are R₁ = 4 Ω and R₂ = 6 Ω

Electrical energy = ?

As the battery and resistors are connected in series, equivalent resistance is given by

R = 2 + 4 + 6 = 12 Ω

From Ohm's law —

V = IR

Substituting the values in the formula above, we get,

15 = I x 12
⇒ I = 15 / 12 = 1.25 A

Hence, current in circuit = 1.25 A

Now,

Voltage across 6 Ω = IR = 1.25 × 6 = 7.50 V

Hence, Voltage across 6 Ω = 7.5 V

Time (t) = 1 min = 60 sec

We know, E = $\dfrac{V^2 \times t}{R}$

Substituting the values in the formula above, we get,

$E = \dfrac{7.5^2 \times 60}{6} \\[0.5em] E = 7.5^2 \times 10 \\[0.5em] \Rightarrow E = 562.5 J$

Hence, energy spent across the 6 Ω resistor = 562.5 J