Question

two coil of resistances 3 ohm and 6 ohm are connected in series across a battery of emf 12 Volt. find the electrical energy consumed in 1 minute in each resistance when these are connected in series.

Answer 1

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Answer 2

Answer: The correct answer is 960 kWh.

Explanation:

The expression for the equivalent resistance in series combination is as follows;

$R=R_{1}+R_{2}$

Here, $R_{1}$ and $R_{2}$ are the resistances.

Put $R_{1}= 3 ohm$ and $R_{1}= 6 ohm$.

$R=3 ohm+6 ohm$

$R=9 ohm$

The expression for the power in terms of voltage and resistance is as follows;

$P=\frac{V^{2}}{R}$

Put V= 12 V  and R= 9 ohm.

$P=\frac{(12)^{2}}{9}$

$P=16 W$

The expression for the electrical energy in terms of power and time is as follows;

E=Pt

Here, P is the power and t is the time.

Convert the given time into hour by dividing time 1 min by 60.

Put P= 16 W and t =1 min.

E=(16)(1\60)

E=(16)(1\60)

E= 960 kWh

Therefore, the electrical energy is 960 kWh.