a battery of e.m.f 15 V and internal resistance 2 ohm is connected to two resistors of resistance 4 ohm and 6 ohm joined (a) in series, (b) in parallel. find in each case the electrical energy spent per minute in 6 ohm resistor.

EMF of battery, E = 15 V

Internal resistance of battery, r = 2 Ω

(a)

When 4 Ω and 6 Ω are connected in series with the battery, the resistances including the internal resistance are all in series with the battery.

The equivalent resistance of the circuit is = 4 + 6 + 2 = 12 Ω

So, the current in the circuit is = 15/12 = 5/4 A

The energy spent per min (60 s) in 6 Ω resistance is = (5/4)² × 6 × 60 J

= 562.5 J

(b)

When 4 Ω and 6 Ω are joined in parallel with the battery, then the circuit would look like:

The equivalent resistance of 4 Ω and 6 Ω is = (4)(6)/(4+6) = 2.4 Ω

The equivalent resistance of the circuit is = 2.4 + 2 = 4.4 Ω

So, the current drawn from the battery is = 15/4.4 A

Now,

Voltage drop across 6 Ω + voltage drop across 2 Ω = 15 V

Thus, the voltage drop across 6 Ω is = 15 – 2 × (15/4.4) = 8.2 V

So, energy spent in 6 Ω in 1 min or 60 s is = (8.2²/6) × 60 = 672.4 J